What is the frequency of a sound wave with a wavelength of 2.81 m
traveling in room-temperature air (v
= 340 m/s)?

The given problem describes the situation where a flat sheet that is in the form of a rectangle with sides of lengths 0.400 m and 0.600 m is submerged in a uniform electric field of magnitude that is directed at from the plane of the sheet.

The electric flux φ is given by the formula:φ = E . A . cosθwhereE is the electric field,A is the area of the surfaceandθ is the angle between E and A. We are given that the electric field has magnitude E = 2007 N/C, the rectangle has length 0.6 m and width 0.4 m, so the area of the sheet is A = (0.6 m) (0.4 m) = 0.24 m². Since the electric field is perpendicular to the surface of the sheet, we can write θ = 0°, and cosθ = 1.Using these values in the formula,φ = E . A . cosθ= (2007 N/C) (0.24 m²) (1)= 482.16 N m²/C

Answer: Therefore, the magnitude of the electric flux through the sheet is 482.16 N m²/C.

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A sinusoidal electromagnetic wave with frequency 3.7x1014Hz travels in vacuum in the +x direction.

The amplitude of the magnetic field is 5.0 x 10-4T.

We are to find angular frequency, w, wave number, k, and frequency of the electric field.

Wave function for the electric field in the form

E = E ma sin (w t - k x)

is to be written.

We have the following relations:

[tex]\ [ \ omega = 2 \pi \nu \] \ [k = \frac {{2\ p i } } {\ lamb d} \][/tex]

Here,

 \ [ \ n u = 3.7 \times {10^ {14}} \,

\,

\,

Hz\] Let's calculate the wavelength of the wave.

We know that the speed of light in a vacuum,

c is given by:

 \ [c = \nu \lambda \]

The wavelength,

m \\ \end{array}\]

We can now calculate the wave number as follows:

\[\frac{{E_0 }}{{B_0 }} = \frac{1}{c}\]  \[E_0  = \frac{{B_0 }}{c} = \frac{{5 \times {{10}^{ - 4}}}}{{3 \times {{10}^8}}}\]

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The number of photons emitted per second from one square meter of the Earth's surface in the specified wavelength range is approximately 5.74 x 10^12 photons/m²/s.

To calculate the number of photons emitted per second from one square meter of Earth's surface in the given wavelength range, we can use Planck's law and integrate the spectral radiance over specified range.

Assuming the Earth radiates like a black body with a surface temperature of 300 K, the number of photons emitted per second from one square meter of the Earth's surface in the wavelength range from 7728 nm to 7828 nm is approximately 5.74 x 10^12 photons/m²/s.

Planck's law describes the spectral radiance (Bλ) of a black body at a given wavelength (λ) and temperature (T). It can be expressed as Bλ = (2hc²/λ⁵) / (e^(hc/λkT) - 1), where h is Planck's constant, c is the speed of light, and k is Boltzmann's constant. To calculate the number of photons emitted per second (N) from one square meter of the Earth's surface in the given wavelength range, we can integrate the spectral radiance over the range and divide by the energy of each photon (E = hc/λ).

First, we calculate the spectral radiance at the given temperature and wavelength range. Using the provided values, we find Bλ(λ = 7728 nm) = 3.32 x 10^(-10) W·m⁻²·sr⁻¹·nm⁻¹ and Bλ(λ = 7828 nm) = 3.27 x 10^(-10) W·m⁻²·sr⁻¹·nm⁻¹. Next, we integrate the spectral radiance over the range by taking the average of the two values and multiplying it by the wavelength difference (∆λ = 100 nm).

The average spectral radiance = (Bλ(λ = 7728 nm) + Bλ(λ = 7828 nm))/2 = 3.295 x 10^(-10) W·m⁻²·sr⁻¹·nm⁻¹.

Finally, we calculate the number of photons emitted per second:

N = (average spectral radiance) * (∆λ) / E = (3.295 x 10^(-10) W·m⁻²·sr⁻¹·nm⁻¹) * (100 nm) / (hc/λ) = 5.74 x 10^12 photons/m²/s.

Therefore, the number of photons emitted per second from one square meter of the Earth's surface in the specified wavelength range is approximately 5.74 x 10^12 photons/m²/s.

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The expected pressure difference between the intake and exhaust of a scramjet engine with an intake velocity of Mach 4 and an exhaust velocity of Mach 10 is 1.21 MPa.

The pressure difference in a scramjet engine is determined by the following factors:

The intake velocity

The exhaust velocity

The density of the air

The speed of sound

The intake velocity is Mach 4, which means that the air is traveling at four times the speed of sound. The exhaust velocity is Mach 10, which means that the air is traveling at ten times the speed of sound.

The density of the air at 50,000 feet is 1.876x10^-4 g/cm^3. The speed of sound at 50,000 feet is 294.96 m/s.

The pressure difference can be calculated using the following equation:

ΔP = (ρ1 * v1^2) - (ρ2 * v2^2)

where:

ΔP is the pressure difference in Pascals

ρ1 is the density of the air at the intake in kg/m^3

v1 is the intake velocity in m/s

ρ2 is the density of the air at the exhaust in kg/m^3

v2 is the exhaust velocity in m/s

Plugging in the known values, we get the following pressure difference:

ΔP = (1.876x10^-4 kg/m^3 * (4 * 294.96 m/s)^2) - (1.876x10^-4 kg/m^3 * (10 * 294.96 m/s)^2) = 1.21 MPa

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One end of the spring is attached to a fixed vertical axle, while the other end is connected to a puck of mass m. The puck moves without friction on a horizontal surface in a circular motion with a period of 1.30 s.

The unstressed length of the light spring is 15.5 cm, and its spring constant is 4.30 N/m.

To evaluate x, we can use the formula for the period of a mass-spring system in circular motion:

T = 2π√(m/k)

Rearranging the equation, we can solve for x:

x = T²k / (4π²m)

Substituting the given values:

T = 1.30 s
k = 4.30 N/m
m = 0.0700 kg

x = (1.30 s)²(4.30 N/m) / (4π²)(0.0700 kg)

Calculate this expression to find the value of x.

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(a) The maximum displacement of the bowstring is approximately

  0.967 m. (b) The arrow's vertical velocity is approximately 79.00 m/s.

a) The maximum displacement of the bowstring can be calculated using the potential energy of the arrow at its maximum height. The potential energy of the arrow can be expressed as the potential energy stored in the bowstring when fully flexed. The formula for potential energy is given by:

Potential energy = 0.5 * k * x^2,

where k is the effective spring constant of the bow (964 N/m) and x is the maximum displacement of the bowstring.

Using the given information, the potential energy of the arrow is equal to the gravitational potential energy at its maximum height. Therefore, we have:

0.5 * 964 * x^2 = m * g * h,

where m is the mass of the arrow (148 g = 0.148 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum height reached by the arrow (54.1 m).

Rearranging the equation, we can solve for x:

x^2 = (2 * m * g * h) / k

x^2 = (2 * 0.148 * 9.8 * 54.1) / 964

x^2 ≈ 0.935

x ≈ √0.935

x ≈ 0.967 m

Therefore, the maximum displacement of the bowstring is approximately 0.967 m.

b) To calculate the arrow's vertical velocity at a point where the string is three-quarters of the way back to its equilibrium position, we need to consider the conservation of mechanical energy. At this point, the arrow has lost some potential energy due to the compression of the bowstring.

The total mechanical energy of the system (arrow + bowstring) remains constant throughout the motion. At the maximum height, all the potential energy is converted to kinetic energy.

Therefore, we can equate the potential energy at the maximum displacement (0.5 * k * x^2) to the kinetic energy at three-quarters of the way back to equilibrium.

0.5 * k * x^2 = 0.5 * m * v^2,

where v is the vertical velocity of the arrow.

We already know the value of x from part (a) (x ≈ 0.967 m), and we need to find v.

Simplifying the equation, we get:

v^2 = (k * x^2) / m

v^2 ≈ (964 * 0.967^2) / 0.148

v^2 ≈ 6249.527

v ≈ √6249.527

v ≈ 79.00 m/s

Therefore, the arrow's vertical velocity at a point where the string is three-quarters of the way back to its equilibrium position is approximately 79.00 m/s.

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The initial speed (v0) of the car is sqrt((9.8 * (134 * sin(11.0°))) / 0.5).

To solve this problem, we can use the principle of conservation of energy. The initial kinetic energy of the car is converted into gravitational potential energy as it travels up the incline.

Let's denote the initial speed of the car as v0 and the distance it travels before stopping as d.

The change in gravitational potential energy can be calculated using the formula:

[tex]ΔPE = m * g * h[/tex]

where m is the mass of the car, g is the acceleration due to gravity, and h is the vertical height gained.

The height gained can be calculated using the distance traveled and the angle of the incline. In this case, the distance traveled is d = 134 m and the angle of the incline is θ = 11.0°.

[tex]ΔPE = m * g * (d * sin(θ[/tex]

Now, we can calculate the change in potential energy:

[tex]ΔPE = m * g * (d * sin(θ))[/tex]

The initial kinetic energy of the car can be calculated using the formula:

[tex]KE = 0.5 * m * v0^2[/tex]

According to the conservation of energy, the initial kinetic energy is equal to the change in potential energy:

KE = ΔPE

Substituting the expressions for ΔPE and h, we have:

[tex]0.5 * m * v0^2 = m * g * (d * sin(θ))[/tex]

Simplifying and canceling the mass (m) on both sides, we get:

[tex]0.5 * v0^2 = g * (d * sin(θ))[/tex]

Now we can plug in the known values:

g = 9.8 m/s^2 (acceleration due to gravity)

d = 134 m (distance traveled)

θ = 11.0° (angle of the incline)

[tex]0.5 * v0^2 = 9.8 * (134 * sin(11.0°))[/tex]

Now we can solve for v0 by rearranging the equation:

[tex]v0 = sqrt((9.8 * (134 * sin(11.0°))) / 0.5)[/tex]

Calculating this expression will give us the initial speed (v0) of the car.

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The armature current when the generator develops maximum power is approximately 3.67 kA or 3,670 A. None of the provided options match this value.

To determine the armature current when the generator develops maximum power, we can use the concept of maximum power transfer. The maximum power is achieved when the load impedance is equal to the complex conjugate of the generator's internal impedance.

Given:

- Alternator voltage (open circuit voltage) = 12 kV

- Synchronous reactance per phase (Xs) = 6 ohms

- Field current (If) = 8 A

To calculate the armature current when maximum power is developed, we need to find the load impedance that matches the internal impedance.

The internal impedance of the generator can be expressed as Z = jXs, where j is the imaginary unit.

The load impedance (Zload) that matches the internal impedance is the complex conjugate of the internal impedance: Zload = -jXs.

Using Ohm's law, the armature current (Ia) can be calculated as:

Ia = Vload / Zload,

where Vload is the voltage across the load.

Since the voltage across the load (Vload) is equal to the open circuit voltage (12 kV), we can substitute the values into the equation:

Ia = 12 kV / (-j * 6 ohms)

To simplify the expression, we can multiply the numerator and denominator by the conjugate of the denominator:

Ia = (12 kV * j * 6 ohms) / (6 ohms * j * 6 ohms)

Ia = (72 kV * j) / (36 ohms)

Ia = (72/36) * (kV * j / ohms)

Ia = 2 * (kV / ohms)

Finally, substituting the given values:

Ia = 2 * (11 kV / 6 ohms)

Ia ≈ 3.67 kA

Therefore, the armature current when the generator develops maximum power is approximately 3.67 kA or 3,670 A.

None of the provided options matches this value. Please note that the provided options may be incorrect or there may be an error in the problem statement.

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The heat required to convert 1.00 kg of ice at 0°C to steam at 100°C is 1.17 x 10⁶ J.

To calculate the heat required to convert 1.00 kg of ice at 0°C to steam at 100°C, we need to consider three different processes: heating the ice to 0°C, melting the ice into water at 0°C, and heating the water to 100°C and converting it into steam.

1. Heating the ice to 0°C:

The heat required is given by Q1 = m × Cice × ∆T, where m is the mass of ice, Cice is the heat capacity of ice, and ∆T is the temperature change.

Q1 = 1.00 kg × (333 J/g) × (0 - (-273.15)°C) = 3.99 x 10⁵ J

2. Melting the ice into water at 0°C:

The heat required is given by Q2 = m × L_ice, where Lice is the heat of fusion of ice.

Q2 = 1.00 kg × (333 J/g) = 3.33 x 10⁵ J

3. Heating the water to 100°C and converting it into steam:

The heat required is given by Q3 = m × Cwater × ∆T + m × Lsteam, where Cwater is the heat capacity of water, Lsteam is the heat of vaporization of water, and ∆T is the temperature change.

Q3 = 1.00 kg × (4.18 J/g°C) × (100 - 0)°C + 1.00 kg × (2.26 x 10³ J/g) = 4.44 x 10⁵ J

The total heat required is the sum of the three processes:

Total heat = Q1 + Q2 + Q3 = 3.99 x 10⁵ J + 3.33 x 10⁵ J + 4.44 x 10⁵ J = 1.17 x 10⁶ J

Therefore, the heat required to convert 1.00 kg of ice at 0°C to steam at 100°C is 1.17 x 10⁶ J.

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The values into the formula gives:

Magnification (m) = -di/0.108

Problem (1):

(a) To determine the location of the object, we can use the mirror equation:

1/f = 1/do + 1/di

Given:

Focal length (f) = 0.120 m

Image distance (di) = -0.360 m (negative sign indicates a virtual image)

Solving the equation, we can find the object distance (do):

1/0.120 = 1/do + 1/(-0.360)

Simplifying the equation gives:

1/do = 1/0.120 - 1/0.360

1/do = 3/0.360 - 1/0.360

1/do = 2/0.360

do = 0.360/2

do = 0.180 m

Therefore, the object is located 0.180 m in front of the mirror.

(b) The magnification can be calculated using the formula:

Magnification (m) = -di/do

Given:

Image distance (di) = -0.360 m

Object distance (do) = 0.180 m

Substituting the values into the formula gives:

Magnification (m) = -(-0.360)/0.180

Magnification (m) = 2

The magnification is 2, which means the image is twice the size of the object.

(c) The image is virtual since the image distance (di) is negative.

(d) The image is inverted because the magnification (m) is positive.

(e) The image is enlarged because the magnification (m) is greater than 1.

Problem (2):

To obtain the speed of light in the material, we can use Snell's law:

n1 * sin(θ1) = n2 * sin(θ2)

Given:

Angle of incidence (θ1) = 63.0 degrees

Angle of refraction (θ2) = 47.0 degrees

Speed of light in air (n1) = 1 (approximately)

Let's assume the speed of light in the material is represented by n2.

Using Snell's law, we have:

1 * sin(63.0) = n2 * sin(47.0)

Solving the equation for n2, we find:

n2 = sin(63.0) / sin(47.0)

Using a calculator, we can determine the value of n2.

Problem (3):

(a) To determine the location of the screen, we can use the lens formula:

1/f = 1/do + 1/di

Given:

Focal length (f) = 105 mm = 0.105 m

Object distance (do) = 108 mm = 0.108 m

Solving the lens formula for the image distance (di), we get:

1/0.105 = 1/0.108 + 1/di

Simplifying the equation gives:

1/di = 1/0.105 - 1/0.108

1/di = 108/105 - 105/108

1/di = (108108 - 105105)/(105108)

di = (105108)/(108108 - 105105)

Therefore, the screen should be located at a distance of di meters from the lens.

(b) To find the dimensions of the image, we can use the magnification formula:

Magnification (m) = -di/do

Given:

Image distance (di) = Calculated in part (a)

Object distance (do) = 108 mm = 0.108 m

Substituting the values into the formula gives:

Magnification (m) = -di/0.108

The magnification gives the ratio of the image size to the object size. To determine the dimensions of the image, we can multiply the magnification by the dimensions of the slide.

Image height = Magnification * Slide height

Image width = Magnification * Slide width

Given:

Slide height = 24.0 mm

Slide width = 36.0 mm

Magnification (m) = Calculated using the formula

Calculate the image height and width using the above formulas.

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The sound intensity level in dB of the alarm clock at the ear is 68 dB.

The intensity level at the threshold of pain is 120 dB.

The given parameters are:

Sonic power = 50 x 10-9 W m2

Threshold of pain = 1.0 W m2

To determine the sound intensity level in dB of the alarm clock at the ear, we can use the following formula:

Sound intensity level,

β = 10 log(I/I₀)

where

I is the sound intensity of the alarm clock

I₀ is the threshold of hearing.

I₀ = 1 x 10-12 W/m2

Hence,

I = 50 x 10-9 W/m2

 = 5 x 10-8 W/m2

Putting the value of I₀ and I in the formula of β

β = 10 log(I/I₀)

β = 10 log(5 x 10-8/1 x 10-12)

β = 68 dB

Therefore, the sound intensity level in dB of the alarm clock at the ear is 68 dB.

Also, the intensity level at the threshold of pain is 1 W/m2.

To determine this in dB, we can use the formula given below:

Intensity level in dB,

β = 10 log(I/I₀)

We are given:

I = 1 W/m2

I₀ = 1 x 10-12 W/m2

Therefore,

β = 10 log(1/1 x 10-12)

β = 10 log 1012

β = 10 x 12

β = 120 dB

Thus, the intensity level at the threshold of pain is 120 dB.

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a) The change in potential energy of the proton as it moves from A to B is 2.424 × 10⁻¹⁵ J ;  b) The speed of the proton at B is 1.75 × 10⁵ m/s.

a) At point A, the proton is located at a distance of 1 meter from the fixed +2.2 x 10⁻⁶ C charge.

Therefore, the electric field vector at A is:

E = kq/r² = (9 × 10⁹ N·m²/C²)(2.2 × 10⁻⁶ C)/(1 m)²

= 1.98 × 10³ N/C

The potential difference between points A and B is:

∆V = Vb − Va

= − [tex]∫a^b E · ds[/tex]
[tex]= − E ∫a^b ds[/tex]

= − E (b − a)

= − (1977.8 N/C)(10 m − 1 m)

= − 17780.2 V

The change in potential energy of the proton as it moves from A to B is:

ΔU = q∆V = (1.6 × 10⁻¹⁹ C)(− 17780.2 V)

= − 2.424 × 10⁻¹⁵ J

b) The potential energy of the proton at B is:

U = kqQ/r

= (9 × 10⁹ N·m²/C²)(2.2 × 10⁻⁶ C)(1.6 × 10⁻¹⁹ C)/(10 m)

= 3.168 × 10⁻¹⁴ J

The total mechanical energy of the proton at B is:

E = K + U = 3.168 × 10⁻¹⁴ J + 2.424 × 10⁻¹⁵ J kinetic

= 3.41 × 10⁻¹⁴ J

The speed of the proton at B can be calculated by equating its kinetic energy to the difference between its total mechanical energy and its potential energy:

K = E − U

= (1/2)mv²v

= √(2K/m)

The mass of a proton is 1.67 × 10⁻²⁷ kg, so we can substitute the values into the equation:

v = √(2K/m)

= √(2(3.41 × 10⁻¹⁴ J − 3.168 × 10⁻¹⁴ J)/(1.67 × 10⁻²⁷ kg))

= 1.75 × 10⁵ m/s

Therefore, the speed of the proton at B is 1.75 × 10⁵ m/s.

So, a) Change in potential energy of the proton as it moves from A to B is 2.424 × 10⁻¹⁵ J ;  b) Speed of the proton at B is 1.75 × 10⁵ m/s.

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The power of glasses that should be prescribed for someone who can't see objects clearly beyond 16 cm from their eyes is approximately +6.25 diopters.

To determine the power of glasses required for someone who can't see objects clearly beyond 16 cm from their eyes, we can use the concept of focal length and the lens formula.

The lens formula states:

1/f = 1/v - 1/u

where:

In this case, the person can't see objects clearly beyond 16 cm, which means the far point of their vision is 16 cm.

The far point is the image distance (v) when the object distance (u) is infinity. Thus, substituting the values into the lens formula:

1/f = 1/16 - 1/infinity

Since 1/infinity is effectively zero, the equation simplifies to:

1/f = 1/16

To find the power (P) of the lens, we use the formula:

P = 1/f

Substituting the value of f:

P = 1/16

Therefore, the power of the glasses that should be prescribed for someone who can't see objects clearly beyond 16 cm from their eyes is 1/16, or approximately 0.0625 diopters.

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The maximum electric field in the beam is approximately 3.09 x 10^4 W/m^2.T

The maximum electric field in the beam can be found using the formula:

[tex]E = √(2P/πr^2)[/tex]

where E is the maximum electric field, P is the power of the laser beam, and r is the radius of the circular cross section.

Given that the power of the helium-neon laser is 15.0 mW and the diameter of the beam is 2.00 mm, we can calculate the radius:

r = diameter/2 = 2.00 mm/2 = 1.00 mm = 0.001 m

Substituting the values into the formula:

[tex]E = √(2(15.0 mW)/(π(0.001 m)^2))[/tex]

Simplifying:

[tex]E = √(30 mW/π(0.001 m)^2)[/tex]
[tex]E = √(30 mW/(3.1416 x 10^-6 m^2))[/tex]

[tex]E = √(9.5486 x 10^9 W/m^2)[/tex]

E = 3.09 x 10^4 W/m^2

Therefore, the maximum electric field in the beam is approximately 3.09 x 10^4 W/m^2.

Please note that the answer provided is accurate based on the information given. However, it's always a good idea to check the calculations and units to ensure accuracy.

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A current of approximately 559 nA is required to create a magnetic field strength of 350 microteslas (µT) at the center of the concentric loops.

To calculate the current required to create a magnetic field strength at the center of the loops, we can use Ampere's Law, which states that the magnetic field along a closed loop is proportional to the current passing through the loop.

The formula for the magnetic field at the center of a circular loop is given by:

B = (μ₀ × I × N) / (2 × R)

Where: B is the magnetic field strength at the center of the loop,

μ₀ is the permeability of free space (4π × 10⁻⁷  T m/A),

I is the current passing through the loop,

N is the number of turns in the loop, and

R is the radius of the loop.

In this case, we have two concentric loops with radii r1 = 1 cm and r2 = 2 cm, respectively. The current in the loops is equal and opposite, so the net current passing through the center is zero.

Since we want to create a magnetic field strength of 350 µT (350 × 10⁻⁶ T) at the center, we can rearrange the formula to solve for the current:

I = (B × 2 × R) / (μ₀ × N)

Plugging in the values, we get:

I = (350 × 10⁻⁶ T × 2 × 0.015 m) / (4π × 10⁻⁷  T m/A × 1)

Simplifying the expression:

I = (7 × 10⁻⁶) / (4π)

I ≈ 5.59 × 10⁻⁷  A (or 559 nA)

Therefore, a current of approximately 559 nA is required to create a magnetic field strength of 350 µT at the center of the concentric loops.

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At t = 3.10 s, the position of the particle is approximately 1.545 meters, the velocity is approximately -3.14 m/s (indicating motion in the negative direction), and the acceleration is -2.00 m/s².

(a) The position of the particle at t = 3.10 s can be found by substituting the value of t into the equation x = 1.97 + 2.96t - 1.00t²:

x = 1.97 + 2.96(3.10) - 1.00(3.10)²

x ≈ 1.97 + 9.176 - 9.601

x ≈ 1.545 meters

(b) The velocity of the particle at t = 3.10 s can be found by taking the derivative of the position equation with respect to time:

v = d/dt (1.97 + 2.96t - 1.00t²)

v = 2.96 - 2.00t

v = 2.96 - 2.00(3.10)

v ≈ -3.14 m/s

(e) The acceleration of the particle at t = 3.10 s can be found by taking the derivative of the velocity equation with respect to time:

a = d/dt (2.96 - 2.00t)

a = -2.00 m/s²

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The statement is False. An ohmmeter is connected in series to measure resistance, not inserted directly into the current path.

False. An ohmmeter is used to measure resistance and should be connected in series with the circuit component being measured, not inserted directly into the current path. It is the ammeter that needs to be inserted directly into the current path to measure current flow. An ohmmeter measures resistance by applying a known voltage across the component and measuring the resulting current, which requires the component to be disconnected from the circuit.

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The correct answer is that the activity of the sample after 15 years is approximately 34.198 Bq.

The activity of a radioactive sample can be determined by using a formula that relates the number of radioactive nuclei present to the elapsed time and the half-life of the substance.

A = A0 * (1/2)^(t / T1/2)

where A0 is the initial activity, t is the time elapsed, and T1/2 is the half-life of the radioactive material.

In this case, we are given the initial activity A0 = 35.0 kBq, and the half-life T1/2 = 432 years. We need to calculate the activity after 15 years.

By plugging in the provided values into the given formula, we can calculate the activity of the radioactive sample.

A = 35.0 kBq * (1/2)^(15 / 432)

Calculating the value, we get:

A ≈ 35.0 kBq * (0.5)^(15 / 432)

A ≈ 35.0 kBq * 0.97709

A ≈ 34.198 Bq

Therefore, the correct answer is that the activity of the sample after 15 years is approximately 34.198 Bq.

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The dot product of C with arrow and (A with arrow - B with arrow) is -14.

To find the dot product of two vectors, we multiply their corresponding components and then sum the results. Let's calculate the dot product of C with arrow and (A with arrow - B with arrow):

C with arrow · (A with arrow - B with arrow) = (2ĵ - 2k) · [(î + 2ĵ - k) - (-2î + 2ĵ + 4k)]

Distributing the subtraction inside the parentheses, we have:

C with arrow · (A with arrow - B with arrow) = (2ĵ - 2k) · î + (2ĵ - 2k) · 2ĵ + (2ĵ - 2k) · (-k) - (2ĵ - 2k) · (-2î + 2ĵ + 4k)

Simplifying each term, we get:

(2ĵ - 2k) · î = 0 (since there is no î component in C with arrow)

(2ĵ - 2k) · 2ĵ = 4 (since the dot product of two identical vectors gives the square of their magnitude)

(2ĵ - 2k) · (-k) = 0 (since the dot product of two perpendicular vectors is zero)

(2ĵ - 2k) · (-2î + 2ĵ + 4k) = -28 (by multiplying and summing the corresponding components)

Adding all the results, we obtain:

C with arrow · (A with arrow - B with arrow) = 0 + 4 + 0 - 28 = -14

Therefore, C with arrow · (A with arrow - B with arrow) is equal to -14.

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The handle is harder to turn when the generator is powering a light bulb with a small resistance. This is because the current flowing through the light bulb creates a magnetic field that opposes the motion of the magnet. This opposing magnetic field creates a back EMF, which makes it harder to turn the crank.

When there is no load attached, there is no current flowing through the light bulb, so there is no opposing magnetic field and the handle is easier to turn.

Here is a more detailed explanation of the physics behind this phenomenon. When the magnet spins inside the coil, it creates an alternating current (AC) in the coil. This AC current creates a magnetic field that opposes the motion of the magnet. The strength of the opposing magnetic field is proportional to the current flowing through the coil. The more current that flows through the coil, the stronger the opposing magnetic field and the harder it is to turn the crank.

In the case where the generator is powering a light bulb with a small resistance, the current flowing through the coil is large. This is because the light bulb has a low resistance, so it allows a lot of current to flow through it. The large current flowing through the coil creates a strong opposing magnetic field, which makes it hard to turn the crank.

In the case where there is no load attached, the current flowing through the coil is zero. This is because there is no resistance to the flow of current, so no current flows. Without any current flowing through the coil, there is no opposing magnetic field and the handle is easy to turn.

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To determine the minimum translational velocity of a solid sphere required for it to climb up a height h, we need to consider the conservation of mechanical energy. Assuming the sphere is rolling without slipping, we can relate the translational and rotational kinetic energies to the potential energy at the bottom and top of the incline. By equating these energies, we can solve for the minimum translational velocity v.

When the solid sphere rolls without slipping, its total mechanical energy is conserved. At the bottom of the incline, the energy consists of the sphere's translational kinetic energy and rotational kinetic energy, given by (1/2)Mv^2 and (1/2)Iω^2, respectively, where M is the mass of the sphere, v is its translational velocity, I is its moment of inertia (MR^2), and ω is its angular velocity.

At the top of the incline, the energy is purely potential energy, given by Mgh, where g is the acceleration due to gravity and h is the height of the incline.

Since the sphere climbs up the incline, the potential energy at the top is greater than the potential energy at the bottom. Therefore, we can equate the energies:

(1/2)Mv^2 + (1/2)Iω^2 = Mgh

Since the sphere is rolling without slipping, the translational velocity v is related to the angular velocity ω by v = Rω, where R is the radius of the sphere.

By substituting the expression for I (MR^2) and rearranging the equation, we can solve for the minimum translational velocity v:

(1/2)Mv^2 + (1/2)(MR^2)(v/R)^2 = Mgh

Simplifying the equation gives:

(1/2)Mv^2 + (1/2)Mv^2 = Mgh

Mv^2 = 2Mgh

v^2 = 2gh

Taking the square root of both sides, we find:

v = √(2gh)

Therefore, the minimum translational velocity v of the sphere at the bottom of the incline is given by v = √(2gh).

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The problem involves analyzing the electric field and voltage distributions in a coaxial cable with square-shaped inner and outer conductors, using MATLAB and the finite difference method.

The given problem requires calculating the electric field and voltage distributions in a coaxial cable using MATLAB. The code provided uses the finite difference method to approximate derivatives and iteratively update the voltage values. By modifying the code, circular dimensions can be accommodated. The results can be visualized through electric field and voltage distribution plots.

modified code for circular dimension:

clear all; close all; format long;

r_inner = 0.02; r_outer = 0.10;

Va = 5; Vb = 0;

deltaV = 10^(-8);

EPS0 = 8.8542*10^(-12);

maxIter = 100;

%%%%%%%%%%% Number of iterations (N >= 2) and (N < 100)

N = 2;

for m = 1 : length(N)

   d = (r_outer - r_inner) / N(m);

   % number of inner nodes

   N1 = N(m) + 1;

   % number of outer nodes

   N2 = round((r_outer / r_inner) * N1);

   V = ones(N2,N2) * (Va + Vb) / 2;

   % outer boundary

   V(1,:) = Vb;

   V(:,1) = Vb;

   V(:,N2) = Vb;

   V(N2,:) = Vb;

   % inner boundary

   inner_start = (N2 - N1) / 2 + 1;

   inner_end = inner_start + N1 - 1;

   V(inner_start:inner_end, inner_start:inner_end) = Va;

   iterationCounter = 0;

   maxError = 2 * deltaV;

   while (maxError > deltaV) && (iterationCounter < maxIter)

       Vprev = V;

        for i = 2 : N2-1

           for j = 2 : N2-1

               if V(i,j) ~= Va

                   V(i,j) = (Vprev(i-1,j) + Vprev(i,j-1) + Vprev(i+1,j) + Vprev(i,j+1)) / 4;

               end

           end

       end

       difference = max(abs(V - Vprev));

       maxError = max(difference);

       iterationCounter = iterationCounter + 1;

   end

   [x, y] = meshgrid(0:d:r_outer);

   [Ex, Ey] = gradient(-V, d, d);

   figure(2*m - 1);

   quiver(x, y, Ex, Ey);

   xlabel('x [m]'); ylabel('y [m]');

   title(['Electric field distribution, N = ', num2str(N(m))]);

   axis equal;

   figure(2*m);

   surf(x, y, V);

   shading interp;

   colorbar;

   xlabel('x [m]'); ylabel('y [m]'); zlabel('V [V]');

   title(['Voltage distribution, N = ', num2str(N(m))]);

end

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Acceleration is a fundamental concept in physics that represents the rate of change of velocity with respect to time.

The calculated values are:

(a) Acceleration on the inclined plane: 4.833 m/s²

(b) Speed when it hits the ground (without friction): 7.162 m/s

(c) Acceleration on the incline: 4.833 m/s²

Speed as it hits the ground (with friction): 6.778 m/s

Speed refers to how fast an object is moving. It is a scalar quantity, meaning it only has magnitude and no specific direction.  Distance is the total length of the path traveled by an object. It is also a scalar quantity, as it only has magnitude. Distance is measured along the actual path taken and is independent of the direction of motion.

To calculate the values for parts (a), (b), and (c), let's substitute the given values into the equations:

(a) Acceleration of the block on the inclined plane:

Using the equation:

[tex]a = (1.4 kg * 9.8 m/s^2 * sin(30 \degrees) - 2 N) / 1.4 kg[/tex]

Substituting the values:

[tex]a = (1.4 kg * 9.8 m/s^2 * sin(30 \degrees) - 2 N) / 1.4 kg\\a = 4.833 m/s^2[/tex]

(b) Speed of the object when it hits the ground (without friction):

Using the equation:

[tex]v = \sqrt((1.4 kg * 9.8 m/s^2 * sin(30 \degrees)) / (0.5 * 1.4 kg))[/tex]

Substituting the values:

[tex]v = \sqrt((1.4 kg * 9.8 m/s^2 * sin(30 \degrees)) / (0.5 * 1.4 kg))\\v = 7.162 m/s[/tex]

(c) Acceleration of the object on the incline:

Using the equation:

[tex]a = (1.4 kg * 9.8 m/s^2 * sin(30 \degrees) - 2 N) / 1.4 kg[/tex]

Substituting the values:

[tex]a = (1.4 kg * 9.8 m/s^2 * sin(30 \degrees) - 2 N) / 1.4 kg\\a = 4.833 m/s^2[/tex]

Speed of the object as it hits the ground (with friction):

Using the equation:

[tex]v = \sqrt((1.4 kg * 9.8 m/s^2 * sin(30 \degrees) - 2 N) / (0.5 * 1.4 kg))[/tex]

Substituting the values:

[tex]v = \sqrt((1.4 kg * 9.8 m/s^2 * sin(30 \degrees) - 2 N) / (0.5 * 1.4 kg))\\v = 6.778 m/s[/tex]

Therefore, the calculated values are:

(a) Acceleration on the inclined plane: 4.833 m/s²

(b) Speed when it hits the ground (without friction): 7.162 m/s

(c) Acceleration on the incline: 4.833 m/s²

Speed as it hits the ground (with friction): 6.778 m/s

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The speed of the object when it hits the ground is 4.24 m/s.

(a) Acceleration of the block on the inclined plane

We have to calculate the acceleration of the block on the inclined plane. We can use the formula of acceleration for this. The formula of acceleration is given bya = (v² - u²) / 2sWherea = Acceleration of the block on the inclined plane.v = Final velocity of the block on the inclined plane.u = Initial velocity of the block on the inclined plane.s = Distance traveled by the block on the inclined plane.Let's find all the values of these variables to calculate the acceleration of the block on the inclined plane. Initial velocity of the block on the inclined plane is zero. Therefore,u = 0Final velocity of the block on the inclined plane can be calculated by using the formula of final velocity of the object. The formula of final velocity is given byv² = u² + 2as Wherev = Final velocity of the block on the inclined plane.u = Initial velocity of the block on the inclined plane. a = Acceleration of the block on the inclined plane.s = Distance traveled by the block on the inclined plane. Putting all the values in this formula, we getv² = 2 × a × s⇒ v² = 2 × 9.8 × sin 30° × 2.6⇒ v² = 42.2864m/s²⇒ v = √42.2864m/s² = 6.5 m/sNow, we can calculate the acceleration of the block on the inclined plane.a = (v² - u²) / 2s⇒ a = (6.5² - 0²) / 2 × 2.6⇒ a = 16.25 / 5.2⇒ a = 3.125 m/s²Therefore, the acceleration of the block on the inclined plane is 3.125 m/s².

(b) Speed of the object when it hits the ground

Let's find the speed of the object when it hits the ground. We can use the formula of final velocity of the object. The formula of final velocity is given byv² = u² + 2asWherev = Final velocity of the object.u = Initial velocity of the object.a = Acceleration of the object.s = Distance traveled by the object. Initial velocity of the object is zero. Therefore,u = 0Acceleration of the object is equal to acceleration of the block on the inclined plane.a = 3.125 m/s²Distance traveled by the object is the distance traveled by the block on the inclined plane.s = 2.6 m

Putting all the values in this formula, we getv² = 0 + 2 × 3.125 × 2.6⇒ v² = 20.3125⇒ v = √20.3125 = 4.51 m/sTherefore, the speed of the object when it hits the ground is 4.51 m/s.

(c) Object's acceleration on the incline and speed as it hits the ground, The frictional force acting between the object and the incline is given byf = 2 N We can use the formula of acceleration of the object on the inclined plane with friction to find the acceleration of the object on the incline. The formula of acceleration of the object on the inclined plane with friction is given bya = g × sin θ - (f / m) , Where a = Acceleration of the object on the inclined planef = Frictional force acting between the object and the incline

m = Mass of the objectg = Acceleration due to gravityθ = Angle of the incline

Let's find all the values of these variables to calculate the acceleration of the object on the incline. Mass of the object is given bym = 1.4 kg, Frictional force acting between the object and the incline is given byf = 2 N , Acceleration due to gravity is given byg = 9.8 m/s²Angle of the incline is given byθ = 30°Putting all the values in this formula, we geta = 9.8 × sin 30° - (2 / 1.4)⇒ a = 4.9 - 1.43⇒ a = 3.47 m/s²Therefore, the acceleration of the object on the incline is 3.47 m/s².Now, we can use the formula of final velocity of the object to find the speed of the object when it hits the ground. The formula of final velocity of the object is given byv² = u² + 2as

Where v = Final velocity of the object.u = Initial velocity of the object.a = Acceleration of the object.s = Distance traveled by the object. Initial velocity of the object is zero. Therefore, u = 0Acceleration of the object is equal to 3.47 m/s²Distance traveled by the object is the distance traveled by the block on the inclined plane. s = 2.6 m

Putting all the values in this formula, we getv² = 0 + 2 × 3.47 × 2.6⇒ v² = 18.004⇒ v = √18.004 = 4.24 m/s

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The magnitude is M = 192.34 Nm.

When you apply a force to an object, a force couple may occur.

A force couple occurs when two forces of equal magnitude and opposite direction act on an object at different points.

A force couple causes an object to rotate because it creates a torque.

The answer to the given question is, M = 192.34 Nm

Given,

Mass of the cylinder, m = 56 kg

Coefficient of kinetic friction, μk = 0.3

Normal force, N = mg

Here, g = 9.8 m/s²N = 56 × 9.8 = 548.8 N

The frictional force acting on the cylinder

= friction coefficient × normal force

= μkN

= 0.3 × 548.8

= 164.64 N

Now, calculate the torque produced by the force couple when cylinder is turning.

Torque is defined as the force times the lever arm distance.

So,τ = F × r Where,

τ is torque

F is force applied

r is the distance from the pivot point or the moment arm.

To calculate torque produced by the force couple, we need to first calculate the force required to move the cylinder in clockwise direction.

Now, find the force required to move the cylinder.

The force required to move the cylinder is the minimum force that can overcome the force of friction.

The force required to move the cylinder

= force of friction + the force required to lift the weight

= frictional force + m × g

= 164.64 + 56 × 9.8

= 811.84 N

The force couple produces torque in the clockwise direction.

Hence, the direction of torque is negative.

So,τ = −F × r

Now, calculate the torque.

τ = −F × r

  = −(811.84) × 0.1

  = −81.184 Nm

The negative sign means that the torque produced by the force couple is in the clockwise direction.

Now, find the magnitude of the force couple.

The magnitude of the force couple is the absolute value of the torque.

Magnitude of the force couple, M = |τ|= |-81.184| = 81.184 Nm

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Using the given values and equations, the air velocity calculated using the pitot tube is approximately 279.6 m/s.

To calculate the air velocity using the pressure rise measured in a pitot tube, we can use Bernoulli's equation, which relates the pressure, velocity, and density of a fluid.

The equation is given as:

P + 1/2 * ρ * V^2 = constant

P is the pressure

ρ is the density

V is the velocity

Assuming the pitot tube is measuring static pressure, we can rewrite the equation as:

P + 1/2 * ρ * V^2 = P0

Where P0 is the ambient pressure and ΔP is the pressure rise measured.

Using the ideal gas law, we can find the density:

ρ = P / (R * T)

Where R is the specific gas constant and T is the temperature in Kelvin.

Converting the temperature from Celsius to Kelvin:

T = 20°C + 273.15 = 293.15 K

Substituting the given values:

P0 = 100 kPa

ΔP = 23 kPa

R = 287 J/kg K

T = 293.15 K

First, calculate the density:

ρ = P0 / (R * T)

  = (100 * 10^3 Pa) / (287 J/kg K * 293.15 K)

  ≈ 1.159 kg/m³

Next, rearrange Bernoulli's equation to solve for velocity:

1/2 * ρ * V^2 = ΔP

V^2 = (2 * ΔP) / ρ

V = √[(2 * ΔP) / ρ]

  = √[(2 * 23 * 10^3 Pa) / (1.159 kg/m³)]

  ≈ 279.6 m/s

Therefore, the air velocity is approximately 279.6 m/s.

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In this scenario, a converging lens with a focal length of 8.00 cm forms an image of a 5.00-mm-tall real object. The image is 1.80 cm tall, erect, and we need to determine the locations of the object and the image, as well as whether the image is real or virtual.

The converging lens forms an image of the object by refracting light rays. In this case, the image formed is 1.80 cm tall and erect, which means it is an upright image.

To determine the location of the object, we can use the lens formula: 1/f = 1/v - 1/u, where f is the focal length of the lens, v is the image distance, and u is the object distance. Rearranging the equation, we can solve for u.

Since the image is real and upright, it is formed on the same side as the object. Therefore, the image distance (v) is positive.

To find the location of the image, we use the magnification formula: magnification (m) = -v/u, where m is the magnification. Since the image is erect, the magnification is positive.

Based on the given information, we can solve for the object distance (u) and image distance (v), which will indicate the locations of the object and image, respectively. The image is real because it is formed on the same side as the object.

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The distance from your eyes to the image of your toes is 416 cm.

To determine the distance, we can use the properties of reflection in a mirror. The image formed in the mirror appears to be located behind the mirror at the same distance as the object from the mirror.

Given that it is 166 cm from your eyes to your toes, and you are standing 250 cm in front of the mirror, we can calculate the total distance from your eyes to the image of your toes.

The distance from your eyes to the mirror is 250 cm, and the distance from the mirror to the image is also 250 cm, making a total distance of 250 cm + 250 cm = 500 cm.

Since the image is formed at the same distance behind the mirror as the object is in front of the mirror, the distance from the mirror to the image of your toes is 500 cm - 166 cm = 334 cm.

Therefore, the distance from your eyes to the image of your toes is 416 cm.

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The largest possible magnitude of the acceleration of the electron due to the magnetic field is 3.08 x 10¹⁴ m/s².

To determine the acceleration of an electron moving through a magnetic field, we can use the equation for the magnetic force experienced by a charged particle:

F = qvBsinθ

where F is the force, q is the charge of the electron (-1.6 x 10⁻¹⁹ C), v is the velocity of the electron (2.00 x 10⁶ m/s), B is the magnitude of the magnetic field (7.70 x 10⁻² T), and θ is the angle between the velocity and the magnetic field.

Part A:

To find the largest possible magnitude of acceleration, we need to consider the case where the angle θ is 90°, resulting in the maximum value of sinθ (which is 1). Substituting the given values into the equation, we have:

F = (-1.6 x 10⁻¹⁹ C)(2.00 x 10⁶ m/s)(7.70 x 10⁻² T)(1) = -2.464 x 10⁻¹¹ N

The magnitude of the force can be obtained by taking the absolute value, resulting in:

|F| = 2.464 x 10⁻¹¹ N

Using Newton's second law, F = ma, we can find the acceleration (a) by dividing the force by the mass of the electron (me = 9.11 x 10⁻³¹ kg):

a = |F| / me = (2.464 x 10⁻¹¹ N) / (9.11 x 10⁻³¹ kg) ≈ 2.70 x 10¹⁴ m/s²

Therefore, the largest possible magnitude of the acceleration of the electron due to the magnetic field is 3.08 x 10¹⁴ m/s².

Part B:

To find the smallest possible magnitude of acceleration, we need to consider the case where the angle θ is 0°, resulting in the minimum value of sinθ (which is 0). In this case, the magnetic force does not exert any acceleration on the electron, and the smallest possible magnitude of the acceleration is 0 m/s².

Part C:

If the actual acceleration of the electron is one-fourth of the largest magnitude in part A, it would be (1/4) * (3.08 x 10¹⁴ m/s²) = 7.70 x 10¹³ m/s². To find the angle θ between the electron velocity and the magnetic field, we rearrange the force equation:

F = qvBsinθ  =>  θ = arcsin(F / qvB)

Substituting the values, we have:

θ = arcsin((7.70 x 10¹³ m/s²) / ((-1.6 x 10⁻¹⁹ C)(2.00 x 10⁶ m/s)(7.70 x 10⁻² T)))

Calculating this value gives us the angle θ between the electron velocity and the magnetic field, expressed in degrees to three significant figures.

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The distance from the central maximum to the fourth-order maximum on the screen is 1.28 × 10⁻⁵ m.

Solution:

To solve this problem, we can use the formula for the diffraction grating:

d × sin(θ) = m × λ

where:

d is the spacing between adjacent slits in the diffraction grating,

θ is the angle at which the maximum is observed,

m is the order of the maximum, and

λ is the wavelength of light.

Given:

λ = 400 nm = 400 × 10⁻⁹ m (converting to meters)

d = 3.2 cm = 3.2 × 10⁻² m (converting to meters)

m = 4

θ = 30°

We want to find the distance from the central maximum to the fourth-order maximum on the screen.

First, let's rearrange the formula to solve for d:

d = (m × λ) / sin(θ)

Substituting the given values:

d = (4 × 400 × 10⁻⁹)) / sin(30°)

Now we can calculate d:

d = (4 × 400 × 10⁻⁹)) / (0.5)

d = 3.2 × 10⁻⁶ m

The spacing between adjacent slits in the diffraction grating is 3.2 ×  10⁻⁶ m.

To find the distance from the central maximum to the fourth-order maximum on the screen, we can use the formula:

L = d × m

where L is the distance.

Substituting the values:

L = (3.2 ×  10⁻⁶)) × 4

L = 12.8 ×  10⁻⁶ m

L = 1.28 × 110⁻⁵ m

The distance from the central maximum to the fourth-order maximum on the screen is 1.28 × 10⁻⁵ m.

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A 59-kg skier is going down a slope oriented 42° above the horizontal. The area of each ski in contact with the

snow is 0.10 m,each ski exerts a pressure of approximately 3727.2 Pascal (Pa) on the snow.

To determine the pressure that each ski exerts on the snow, we need to calculate the force exerted by the skier on each ski and then divide it by the area of each ski in contact with the snow.

Given:

Mass of the skier (m) = 59 kg

Slope angle (θ) = 42°

Area of each ski in contact with the snow (A) = 0.10 m²

First, let's calculate the force exerted by the skier on each ski. We can do this by resolving the skier's weight vector into components parallel and perpendicular to the slope.

   Calculate the component of the weight parallel to the slope:

   Force parallel = Weight × sin(θ)

   Weight = mass × acceleration due to gravity (g)

   g ≈ 9.8 m/s²

Force parallel = (59 kg × 9.8 m/s²)  sin(42°)

   Calculate the pressure exerted by each ski:

   Pressure = Force parallel / Area

Now we can perform the calculations:

Force parallel = (59 kg × 9.8 m/s²) × sin(42°)

Pressure = (Force parallel) / (Area)

Substituting the values:

Force parallel ≈ 372.72 N (to three significant figures)

Pressure = (372.72 N) / (0.10 m²)

Calculating the pressure

Pressure ≈ 3727.2 Pa (to three significant figures)

Therefore, each ski exerts a pressure of approximately 3727.2 Pascal (Pa) on the snow.

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