Answer:
v₀ = 677.94 m / s , θ = 286º
Explanation:
We can solve this exercise using the kinematic expressions, let's work on each axis separately.
X axis
has a relation of aₓ = 5.10 m / s², the motor is on for a time of t = 675 s, reaching the speed vₓ = 3630 m / s, let's use the relation
vₓ = v₀ₓ + aₓ t
v₀ₓ = vₓ - aₓ t
let's calculate
v₀ₓ = 3630 - 5.10 675
v₀ₓ = 187.5 m / s
Y Axis
[tex]v_{y}[/tex] = v_{oy} - a_{y} t
v_{oy} = v_{y} - a_{y} t
let's calculate
v_{oy} = 4276 - 7.30 675
v_{oy} = -651.5 m / s
we can give the speed starts in two ways
a) v₀ = (187.5 i ^ - 651.5 j ^) m / s
b) in the form of module and angle
Let's use the Pythagorean theorem
v₀ = [tex]\sqrt{v_{ox}^{2} + v_{oy}^{2} }[/tex]
v₀ = [tex]\sqrt{187.5^{2} +651.5^{2} }[/tex]
v₀ = 677.94 m / s
we use trigonometry
tan θ = [tex]\frac{v_{oy} }{v_{ox} }[/tex]
θ = tan⁻¹ \frac{v_{oy} }{v_{ox} }
θ = tan⁻¹ ([tex]\frac{-651.5}{187.5}[/tex])
θ = -73.94º
This angle measured from the positive side of the x-axis is
θ‘ = 360 - 73.94
θ = 286º
Two identical small charged spheres hang in equilibrium with equal masses (0.02kg). The length of the strings is equal (0.18m) and the angle with the vertical is identical (7^o). Find the magnitude of the charge on each sphere. The acceleration of gravity is 9.8 m/s^2 and the value of Coulomb
Answer:
The value is [tex]q = 3.4 *10^{-6} \ C[/tex]
Explanation:
From the question we are told that
The mass of each sphere is [tex]m_1 = m_2 = m = 0.020 \ kg[/tex]
The length of the string is [tex]l = 0.18 \ m[/tex]
The angle of with the vertical is [tex]\theta = 7^o[/tex]
The acceleration due to gravity is [tex]g = 9.8 \ m/s^2[/tex]
Generally the force acting between the forces is mathematically represented as
[tex]F = T cos \theta = \frac{k* q^2}{ r^2}[/tex]
=> [tex]T cos \theta = \frac{k* q^2}{ r^2}[/tex]
Generally from Pythagoras theorem the radius of the circular curve created by the force is
[tex]r = 2 L sin (\theta )[/tex]
=> [tex]r = 2* 0.180 sin (7)[/tex]
=> [tex]r = 0.043 \ m[/tex]
=> [tex]q = tan \theta * \frac{m * g * r^2 }{k}[/tex]
=> [tex]q = tan(7)* \frac{ 0.02 * 9.8 * 0.043^2 }{9*10^{9}}[/tex]
=> [tex]q = 3.4 *10^{-6} \ C[/tex]
A ball is thrown straight up in the air. For which situations are both the instantaneous velocity and the acceleration zero?
a. on the way up
b. at the top of its flight path
c. on the way down
d. halfway up and halfway down
e. none of the above
Answer: e: none of the above.
Explanation:
For any object in the air, the gravitational acceleration will be always -9.8m/s^2, where the negative sign is because gravity pulls the object down.
The instantaneous velocity, as a function of time, for the case of the ball, is
V(t) = (-9.8m/s^2)*t + v0
Where v0 is the velocity at which the ball is thrown up.
The velocity will be zero when the ball is at the top of its flight pat, in that point the sign of the velocity changes, it stops being positive (so the ball stops going up) and becomes negative (so the ball starts to fall down).
Now, while the instantaneous velocity can be zero during the flight, the acceleration does not, it only will be zero when the object hits the ground. Then the only correct option will be e: none of the above.
(a) A defibrillator passes 12.0 A of current through the torso of a person for 0.0100 s. How much charge moves
Answer:
0.12C
Explanation:
A defibrillator passes 12A through the torso of an individual
The time is 0.0100
The charge can be calculated as follows
= I × t
= 12 × 0.0100
= 0.12C
Hence the charge is 0.12C
A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 10.0 min at 75.0 km/h, 6.0 min at 95.0 km/h, and 45.0 min at 40.0 km/h and spends 40.0 min eating lunch and buying gas.(a) Determine the average speed for the trip.
(b) Determine the distance between the initial and final cities along the route.
Answer:
a) v = 0.515 km / min , b) x_total = 52 km
Explanation:
The measured speed is defined by the distance traveled between the time
v = [tex]\frac{\Delta x}{\Delta t}[/tex]
In this case they give us the speed in several time intervals
let's find the distance traveled in each interval
a) Goes at 75 km/h for 10 min
v = [tex]\frac{x}{t}[/tex]x / t
x₁ = v t
let's reduce speed to km / min
v₁ = 75 km / h (1h / 60 min) = 1.25 km / min
the distance traveled in this time is
x₁ = 1.25 10
x₁ = 12.5 km
b) goes to v = 95 km / h for 6 min
v = 95 km / h (1h 60 min) = 1.5833 km / min
the distance traveled is
x₂ = v₂2 t
x₂ = 1.58333 6
x₂ = 9.5 km
c) goes to v = 40 km / h for 45.0 min
v₃ = 40 km / h (1 h / 60min) = 0.6667 km / min
x₃ = 0.6667 45
x₃ = 30 km
d) t = 40 min, stopped
x₄ = 0
A) let's calculate the average speed of the trip
v =[tex]\frac{x_{1}+x_{2}+x_{3}+x_{4} }{t_{1}+t_{2}+t_{3}+t_{4} }[/tex]
v = (12.5 +9.5 +30 +0) / (10 +6 +45 +40)
v = 52/101
v = 0.515 km / min
B) the distance between the two cities is
x_total = x₁ + x₂ + x₃
x_total = 12.5 +9.5 + 30
x_total = 52 km
Select the correct answer.
What are the units for the spring constant, k?
OA newton meters
OB. newton seconds
OC. newtons/meter
OD. newtons/second
E. newtons/seconds
How much force does it take to accelerate a car mass 1750kg at 20m/s
Answer:
The needed force is 35,000 N.
Explanation:
By the second Newton's law, we know that:
F = M*a
Force equals mass times acceleration.
If we know that the mass of the car is 1750kg, then M = 1750kg, and the acceleration is 20 m/s^2, then a = 20m/s^2
We can replace these values in the original equation to find the value of F.
F = 1750kg*20 m/s^2 = 35,000 N
Does the air in a room or water in a pot have to be at a certain temperature to create convection? Yes or No? and Explain why?
Answer:
Yes
Explanation:
because that temperature creates the bulk movement of molecules due to increase in their kinetic energy
a body of mass 1kg is made to oscillate on a spring of force constant 16 n/m calculate 1 the angular frequency 2 the frequency of oscillation
Explanation:
Given that,
Mass of a body, m = 1 kg
Force constant, k = 16 N/m
We need to find the angular frequency and the frequency of oscillation.
(a) The angular frequency of a body is given by :
[tex]\omega=\sqrt{\dfrac{k}{m}} \\\\=\omega=\sqrt{\dfrac{16}{1}} \\\\=4\ rad/s[/tex]
(b) The frequency of oscillation is given by :
[tex]f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{4}{2\pi}\\\\=\dfrac{2}{\pi}\ Hz[/tex]
Hence, this is the required solution.
A tiny water droplet of radius . descends through air from a high building.
Calculate its terminal velocity. Given that for air = × − −− and density of the water = -3
Correct question is;
A tiny water droplet of radius 0.010 cm descends through air from a high building .Calculate its terminal velocity . Given that η of air = 19 × 10^(-6) kg/m.s and density of water ρ = 1000kg/ms
Answer:
1.146 m/s
Explanation:
We are given;
Radius; r = 0.010 cm = 0.01 × 10^(-2) m
η = 19 × 10^(-6) kg/m.s
ρ = 1000 kg/ms
The formula for the terminal velocity is given by;
V_t = 2r²ρg/9η
g is acceleration due to gravity = 9.8 m/s²
Thus;
V_t = (2 × (0.01 × 10^(-2))² × 1000 × 9.8)/(9 × 19 × 10^(-6))
V_t = 1.146 m/s
Balboa Park in San Diego has an outdoor organ. When the air temperature increases, how does the fundamental frequency of the one of the organ pipes change?
a. increases
b. decreases
c. stays the same
d. is impossible to determine. (The thermal expansion of the pipe is negligible.)
Answer:
a. increases
Explanation:
Outdoor pipe organ is a musical instrument that is used to play music. It produces some soothing notes and is very peaceful to the ears. The organ pipe is the source of sound for the outdoor organ. And the length of the pipe determines the wavelength of the sound. When the air temperature of the surrounding is increased, the fundamental frequency of one of the organ pipes will be increased as the speed of the sound will increased.
What is a
physical
property of
snowflakes?
Plants that respond to light are responding
to an:
a. internal stimulus
b. external stimulus
please answer
Answer:
The answer would be a
Explanation:
A 2.0 kg mass on a horizontal spring is pulled back 2.0 cm and released. If, instead, a 0.40 kg mass were used in this same experiment, the total energy of the system would: ________
a. Remain the same.
b. Double.
c. Be half as large.
Answer:
a. Remain the same.
Explanation:
The total energy of the spring mass system is given as
[tex]$E=\frac{1}{2}kx^2 $[/tex]
where k is the spring constant of the spring
x is the compression or expansion of the spring
The energy of the spring mass system is always independent of mass. It does not depends on the mass object attached to the spring.
Therefore, the total energy of the system will remain the same in the experiment when the mass is changes the second time.
An airplane traveling at half the speed of sound emits a sound of frequency 5.84 kHz. (a) At what frequency does a stationary listener hear the sound as the plane approaches?
Answer:
Stationary listener frequency = 77.68 kHz
Explanation:
Given:
Speed of airplane = 344/2 = 172 m/s
Sound of frequency = 5.84 kHz
Computation:
f0 = fs[(v+v0)/(v-vs)]
f0 = 5.84[(344+0)/(344-172)]
f0 = 5.84[(344)/172]
Stationary listener frequency = 77.68 kHz
PLEASE HELP ME 20 POINTS Some metals have a molecular structure that makes them good conductors. Explain how understanding this relationship can help engineers make more powerful batteries.
PLEASE I KNOW THE STRUCTURE I JUST NEED HOW BY UNDERSTANDING THE RELATIONSHIP THEY CAN MAKE MORE POWERFUL BATTERIES.
PLEASE IM DYEING HERE I JUST NEED THIS ANSWER I WILL GIVE BRAINLIST AND 5 STARS A HEART WHATEVER JUST PLEASEEEEE.
Answer:
negative plus positive with energy = battery
Explanation:
A racing car starts from rest at t=0 and reaches a final speed v at time t. If the acceleration of the car is constant during this time, which of the following statements are true? a) the average speed of the car is v/2; b) the car travels a distance vt; c) the magnitude of the acceleration of the car is v/t; d) the velocity of the car remains constant; or e) none of the statements is true
Answer:
All true statements are shown: a) The average speed of the car is [tex]\frac{v}{2}[/tex], c)The magnitude of the acceleration of the car is [tex]\frac{v}{t}[/tex].
Explanation:
Let prove the validity of each statement:
a) The average speed of the car is [tex]\frac{v}{2}[/tex].
The average speed ([tex]\bar v[/tex]) is defined by the following formula:
[tex]\bar v = \frac{v_{o}+v_{f}}{2}[/tex] (1)
Where:
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the racing car.
If we know that [tex]v_{o} = 0[/tex] and [tex]v_{f} = v[/tex], then the average speed of the racing car:
[tex]\bar v = \frac{0+v}{2}[/tex]
[tex]\bar v = \frac{v}{2}[/tex]
The statement is true.
b) The car travels a distance [tex]v\cdot t[/tex].
Since the racing car is accelerating uniformly, the distance travelled by the car is represented by the following kinematic formula:
[tex]x - x_{o}=v_{o}\cdot t + \frac{1}{2}\cdot a\cdot t^{2}[/tex] (2)
Where [tex]a[/tex] is the acceleration of the racing car, measured in meters per square second.
The statement is false.
c) The magnitude of the acceleration of the car is [tex]\frac{v}{t}[/tex].
Since the racing car is accelerating uniformly, the velocity of the racing car is represented by the following kinematic formula:
[tex]v_{f} = v_{o}+a\cdot t[/tex] (3)
Then, we clear the acceleration of the expression:
[tex]a = \frac{v_{f}-v_{o}}{t}[/tex]
If we know that [tex]v_{o} = 0[/tex] and [tex]v_{f} = v[/tex], then the acceleration of the car is:
[tex]a = \frac{v-0}{t}[/tex]
[tex]a = \frac{v}{t}[/tex]
The statement is true.
d) The velocity of the car remains constant.
Since the car accelerates uniformly, the vehicle does not travel at constant velocity.
The statement is false.
What does a sound wave’s frequency tell you?
a - the number of vibrations per second an object is making
b - the loudness or softness of the sound from an object
c - the amplitude of the sound wave
d - the type of medium in which the sound is traveling
Answer:
The number of vibrations per second an object is making
Explanation:
We need to find the correct option for a sound wave’s frequency.
Frequency is the number of vibrations per second an object is making. It can be calculated as total no. of vibrations divided by time taken.
Its SI unit is Hertz and it is denoted by Hz.
Hence, the correct option is (a).
Nose bleeding occurs as we move towards high altitude, why?
Answer:
Because as we climb higher the amount of oxygen in the air decreases and this makes the air thinner and dryer. And at higher altitudes the blood pressure inside of our body is more than the atmospheric pressure which forces the blood to come out from openings like the nose.
Explanation:
Deduce the first
equation
of motion matematical method
Answer:
The answer is below
Explanation:
Equation of motion is a term in physics that is used in explaining the characteristics of a physical system concerning its motion while considering the period.
The First Equation of Motion is given as v = u + at
Here, a = acceleration, v = final velocity, u = initial velocity, and t = time
To deduce it algebraically, it is written as
a = (v - u) ÷ t
Therefore, it can be repositioned to => v = u + a * t
1) A 40 kg girl gets on her 10 kg wagon on level ground with two 5 kg bricks. She throws the bricks horizontally off the back of the wagon one at a time at a speed of 7 m/s relative to herself. a) How fast does she go after throwing the second brick
Answer:
1.4 m/s
Explanation:
Assuming that the system is an isolated one, then we will conserve the linear momentum of the system so as to be able to solve the system for required quantities:
Using the conservation of momentum, we have,
initial momentum = final momentum
P(i) = P(f)
And from the question, we see that the system has no initial momentum
After throwing the first brick we have
(40 + 10 + 5)v2 + 5 * -7 = 0
55v2 - 35 = 0
55v2 = 35
v2 = 35/55
v2 = 0.63 m/s
Now, considering that she conserves momentum when she throws the second brick again, we then say
(40 + 10 + 5) * 0.63 = (40 + 10)v(f) + 5 * -7
55 * 0.63 = 50v(f) - 35
35 = 50v(f) - 35
50v(f) = 35 + 35
50v(f) = 70
v(f) = 70/50
v(f) = 1.4 m/s
How can music imitate nature?
Answer:
There are at least two ways contemporary music in America has made use of nature by directly making use of the sounds of nature as musical material, or else in works that desire to imitate nature "in its manner of operation." That is, they are constructed the way nature is, they work the way nature works, rather than sounding the way nature sounds.
Explanation:
it takes 10 J of energy to move 2 C of charge from point A to point B. How large is the potential difference between points A and B
Given the workdone and the quantity of the charge, the potential difference between points A and points B is 5 Volts.
What is potential difference?Potential difference is simply referred to as work done required to move a charge from point A to point B per unit charge.
It is expressed as;
V = W / q
Given the data in the question;
Work done W = 10JQuantity of charge q = 2CPotential difference V = ?We substitute our values into the exression above.
V = W / q
V = 10J / 2C
V = 5V
Therefore, given the workdone and the quantity of the charge, the potential difference between points A and points B is 5 Volts.
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A proton with an initial speed of 600,000 m/s is brought to rest by an electric field.1. What was the potential difference that stopped the proton?
2. What was the initial kinetic energy of the proton, in electron volts?
Answer:
(1) the potential difference that stopped the proton is 1878.75 V
(2) the initial kinetic energy of the proton is 1878.75 eV
Explanation:
Given;
initial speed of the proton, v = 600,000 m/s
mass of proton, m = 1.67 x 10⁻²⁷ kg
(1) The work done in bringing the proton to rest is given as;
[tex]W = eV[/tex]
Apply work energy theorem;
[tex]K.E =W\\\\ \frac{1}{2} mv^2 = eV\\\\V = \frac{mv^2}{2e}[/tex]
where;
V is the potential difference
[tex]V = \frac{1.67\times 10^{-27} \times\ 600,000^2}{2 \ \times \ 1.6 \times 10^{-19}} \\\\V = 1878.75 \ V[/tex]
(2) the initial kinetic energy of the proton, in electron volts;
[tex]K.E = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times \ 1.67\times 10^{-27} \times 600,000^2 = 3.006 \times 10^{-16} \ J\\\\K.E = \frac{3.006 \times 10^{-16} \ J \ \ eV}{1.6 \times 10^{-19} \ J} = 1878.75 \ eV[/tex]
How is a mesa formed
Mesas form by weathering and erosion of horizontally layered rocks that have been uplifted by tectonic activity. Variations in the ability of different types of rock to resist weathering and erosion cause the weaker types of rocks to be eroded away, leaving the more resistant types of rocks topographically higher than their surroundings. This process is called differential erosion.
Hope it helps!
Suppose that you take a 10 kg mass on the surface of the earth and then place it on the moon. What will
the mass of the object on the moon be?
Answer:
10 kg
Explanation:
The mass of an object does not change even if the amount of gravtiy changes.
In general, how did the water pressure in the tank change when mass was added to the fluid?
Answer:
As the height increases the pressure must increase.
Explanation:
When we add masses to the fluid, the amount of fluid in the tank increases, therefore its height increases and the pressure is described by the expression
P = ρ g h
where rho is constant for a given fluid and h is the height measured from the surface of the fluid.
As the height increases the pressure must increase.
Which of the following is a correct equation for total energy?
Total energy = Kinetic energy / potential energy
Total energy = Kinetic energy - potential energy
Total energy = Kinetic energy * potential energy
Total energy = Kinetic energy + potential energy
Answer:
kinetic energy + potential energy
Answer:
D is the correct answer: Total energy = Kinetic energy + potential energy
1) Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Chuck, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, Andrea's tangential speed is which of the following?a) twice Chuck's.b) the same as Chuck's.c) half of Chuck's.d) impossible to determine.2) When the merry go round is rotating at a constant angular speed, Andrea's tangential speed is:______.a) twice Chuck's.b) the same as Chuck's.c) half of Chuck's.d) impossible to determine.
Explanation:
The tangential speed of Andrea is given by :
[tex]v=r\omega[/tex]
Where
r is radius of the circular path
ω is angular speed
The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'
r' = 2r
New angular speed,
[tex]v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v[/tex]
New angular speed is twice that of the Chuck's speed.
The 1kg rock is tied to a string and swung in a circular path as shown. The 1 meter string is tied to a post, and during the motion, the string has a 30 angle with the post. The rock makes 100 rounds in 1 minute. The centripetal force on the rock is
Answer: 54.8
Explanation:
The centripetal force on the rock moving along the string is 104.66 N.
The given parameters;
the mass of the rock, m = 1 kgangle of inclination of the string, θ = 30⁰angular velocity of the rock, ω = 100 rev/minlength of the string, r = 1 mThe centripetal acceleration of the rock is calculated as follows;
[tex]\omega _f^2 = \omega _i^2 + 2\alpha (\theta)\\\\(100 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s} )^2 = 0 + 2\alpha (30 ^0 \times \frac{\pi \ rad}{180^0} )\\\\109.69 = 1.048\alpha \\\\\alpha = \frac{109.69}{1.048} \\\\\alpha =104.66 \ rad/s^2[/tex]
[tex]a_c = \alpha \times r\\\\a_c = 104.66 \times 1 = 104.66 \ m/s^2[/tex]
The centripetal force on the rock is calculated as follows;
[tex]F_c = ma_c[/tex]
[tex]F_c = 104.66 \times 1\\\\F_c = 104.66 \ N[/tex]
Thus, the centripetal force on the rock moving along the string is 104.66 N.
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A baseball player slides into home base and experiences coefficients of friction of μ8 = 0.63 and μk = 0.49. The normal force acting on the player is 660N. How much frictional force acts on the player?Round your answer to the nearest hundredth if necessary. Ff = ________ N
Answer:
323.4 N
Explanation:
Given that,
The coefficient of static friction, [tex]\mu_s=0.63[/tex]
The coefficient of kinetic friction, [tex]\mu_k=0.49[/tex]
We need to find the frictional force acting on the player. As the baseball player is sliding into home base, we will use the coefficient of kinetic friction to find the frictional force.
[tex]F=\mu _kN\\\\=0.49\times 660\\\\=323.4\ N[/tex]
So, the frictional force acting on the player is 323.4 N.