Answer:
The resistance of the filament is 333.33 ohmsThe rms current is 0.6 AExplanation:
Given;
output power of the incandescent lightbulb , P = 120 W
input voltage, V = 200 V
The resistance of the filament is calculated as;
[tex]P = \frac{V^2}{R}[/tex]
Where;
R is the resistance of the filament
[tex]R = \frac{V^2}{P} \\\\R = \frac{200^2}{120} \\\\R = 333.33 \ ohms[/tex]
The rms current is given as;
[tex]P = I_{rms} V_{rms}\\\\I_{rms} = \frac{P}{V_{rms}} \\\\I_{rms} =\frac{120}{200}\\\\ I_{rms} = 0.6 \ A[/tex]
The diameter of 0-gauge copper wire is 8.252 mm. Find the resistance of a 1.00-km length of such wire used for power transmission.
Answer:
The value is [tex]R = 0.321 \ \Omega[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 8.252 \ mm = 0.008252 \ m[/tex]
The length of the wire is [tex]l = 1.0 \ km = 1000 \ m[/tex]
Generally the cross sectional area of the copper wire is mathematically represented as
[tex]A = \pi * \frac{d^2}{4}[/tex]
=> [tex]A = 3.142 * \frac{ 0.008252^2}{4}[/tex]
=> [tex]A = 5.349 *10^{ - 5} \ m^2[/tex]
Generally the resistance is mathematically represented as
[tex]R = \frac{\rho * l }{A }[/tex]
Here [tex]\rho[/tex] is the resistivity of copper with the value [tex]\rho = 1.72*10^{-8} \ \Omega \cdot m[/tex]
=> [tex]R = \frac{1.72 *10^{-8} * 1000 }{5.349 *10^{ - 5} }[/tex]
=> [tex]R = 0.321 \ \Omega[/tex]
A child blows a leaf from rest straight up in the air. the leaf has a constant upward acceleration of magnitude 1.0 m by s square. how much time does it take the leaf to displace 1.0m upwards?
Answer:
√2
Explanation:
From the question, we're given that the
Acceleration of the leaf is 1 m/s²
Change in displacement of the leaf is 1 m/s.
Again, from the question, we can tell that the initial velocity u = 0, since the object starts at rest
Now, to solve this, we don't the equation of motion to ur
S = ut + 1/2at², substituting the whole parameters, we then have
1 = 0 * t + 1/2 * 1 * t²
1 = 1/2 * t²
t²/2 = 1
t² = 2
t = √2 seconds
Therefore the time it takes the leaf to dislodge is 2 seconds
In performing rotational motion in an ice capade, which skaters must skate faster, those further in or those further out? Explain the reasoning for your conclusion using concept of rotational inertia.
Answer:
if the angular momentum is constant, the skater is closer, going faster.
w₂ = r₁² / r₂² w₁
Explanation:
For the skaters to perform in the turn in l, the angular number of the same must be equal, so we stammer the concept of conservation of angular momentum
L₁ = L₂
the angular momentum is
L = I w
we substitute
I₁ w₁ = I₂ w₂
w₂ = [tex]\frac{I_{1} }{I_{2} }[/tex] w₁
if we consider the skater as a particle, his moment of inertia is
I = m r²
we substitute
w₂ = r₁² / r₂² w₁
therefore, if the angular momentum is constant, the skater is closer, going faster.
A 0.046 kg golf ball hit by a driver can accelerate from rest to 67 m/s in 1 ms while the driver is in contact with the ball. How much average force does the golf ball experience
Answer:
Average force = 67 mn
Explanation:
Given:
Initial velocity u = 0 m/s
Final velocity v = 67 m/s
Time t = 1 ms = 0.001 sec.
Computation:
Using Momentum theory
Change in momentum = F × Δt
(v-u)/t = F × Δt
F × 0.001 = (67 - 0)/0.001
F= 67,000,000
Average force = 67 mn
How does the spring constant of the smaller springs relate to that of the original spring? Choose the correct explanation.
A) The spring constant of each half will be half the spring constant of the original long spring since it will stretch twice as much under the same tension
B) The spring constant of each half will be twice the spring constant of the original long spring since it will stretch twice as much under the same tension.
C) The spring constant of each half will be twice the spring constant of the original long spring since it will stretch only half as much under the same tension.
D) The spring constant of each half will be half the spring constant of the original long spring since it will stretch only half as much under the same tension.
Answer:
C) The spring constant of each half will be twice the spring constant of the original long spring since it will stretch only half as much under the same tension.
Explanation:
Hooke's law states that the force needed to extend or compress a spring by a distance is proportional to that distance. If is given as:
F = ke, where F is the force applied, k is spring constant and e is the extension.
If a force f is applied to a spring with a spring constant k and by a distance stretched (x) then:
k = F / x
For half the spring, if the same force F is applied, the stretch would be half (x/2), hence the spring constant C is:
C = F / (x/2)
C = 2 (F / x) = 2 * spring constant of original spring
A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of ax = 5.10 m/s2, while the one on the back gives an acceleration component in the ydirection of ay = 7.30 m/s2. The engines turn off after firing for 675 s, at which point the spacecraft has velocity components of vx = 3630 m/s and vy = 4276 m/s.What was the magnitude and direction of the spacecraft's initial velocity, before the engines were turned on?Express the direction as an angle measured counterclockwise from the x -axis.
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
HELPPPPP fasttttt
Do you think an object can travel faster than the speed of sound? Explain your thinking.
Answer: Depends on the context.
Explanation: While simply throwing a football won’t breach the sound barrier, a specialized jet can breach it easily. It all just depends on your definition of an object. Hope this helps!
Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.
Complete Question
Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light
Answer:
The diameter is [tex]D = 0.59 \ m[/tex]
Explanation:
From the question we are told that
The best resolution is [tex]\theta = 0.3 \ arcsecond[/tex]
The wavelength is [tex]\lambda = 700 \ nm = 700 *10^{-9 } \ m[/tex]
Generally the
1 arcminute = > 60 arcseconds
=> x arcminute => 0.3 arcsecond
So
[tex]x = \frac{0.3}{60 }[/tex]
=> [tex]x = 0.005 \ arcminutes[/tex]
Now
60 arcminutes => 1 degree
0.005 arcminutes = > z degrees
=> [tex]z = \frac{0.005}{60 }[/tex]
=> [tex]z = 8.333 *10^{-5} \ degree[/tex]
Converting to radian
[tex]\theta = z = 8.333 *10^{-5} * 0.01745 = 1.454 *10^{-6} \ radian[/tex]
Generally the resolution is mathematically represented as
[tex]\theta = \frac{1.22 * \lambda }{ D}[/tex]
=> [tex]D = \frac{1.22 * \lambda }{\theta }[/tex]
=> [tex]D = \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }[/tex]
=> [tex]D = 0.59 \ m[/tex]
In the attachment there is a density column where there is colour
Question: tell me why is the red at the bottom of the density column if it is the least dense
Answer: The red at the bottom represents the base of the tube, not the red liquid.
Explanation:
The densest materials have more weight per unit of volume.
This means that those elements will always flow to the bottom of the containers, like the one in the image.
The red liquid being the least dense one, can not go to the bottom by its own means.
There could be some cases, like:
The red liquid when solid, is way denser than in its liquid phase, and then the red at the bottom could be solid phase of the red liquid, but there is no mention of this in the question, then we can discard this idea.
Another trivial idea is that the red liquid at the bottom could be trapped by some kind of wall, but again, there is no mention of this, so again we can discard this idea.
The thing that makes sense is that the red at the bottom represents the base of the tube and not the red liquid.
Choose the best answer. A car traveling at constant speed has a net work of zero done on it.
A. True
B. False
Answer:
Explanation:
we know that,
force = mass × acceleration
∴ since speed/velocity is constant, acceleration should be zero.
∴ f = m × 0
f = 0 N
∴ If we apply this to ,
work = force × displacement
we get ,
w = 0 × s
∴ we can say that the net work is zero.
and hence the answer is true!!!
a boulder with a mass of about 1.5 x 10^5 kg falls and strikes the ground at 70 m/s how much kinetic energy dies the boulder deliver to the ground PLEASE HELP
Answer:
k. e= 1/2 mv^2
Ke = 1/2 * 1.5 * 10^5 * 70^2
3.675 *10^8 joules
Two stones, one with twice the mass of the other, are thrown straight up and rise to the same height h. Compare their changes in gravitational potential energy.
a. They rise to the same height, so the stone with twice the mass has twice the change in gravitational potential energy.
b. They rise to the same height, so they have the same change in gravitational potential energy.
c. The answer depends on their speeds at height h.
Answer:
the correct one is the a, U₂ = 2 U₁
Explanation:
The gravitational potential energy is
U = m g h
if the stones reach the same height, the energy of the first stone is
U₁ = m₁ g h
The second stone is twice the mass and reaches the same height
m₂ = 2 m₁
potential energy is
U₂ = m₂ g h
U₂ = 2 m₁ g h
U₂ = 2 U₁
therefore the energy is double.
When reviewing the statements, the correct one is the a
A basketball is rolling rightward onto the court with a speed of 4.0 m/s and slows down with a consta leftward acceleration of magnitude 0.50 m/s^2 over 14 m.
What is the velocity of the basketball after rolling for 14 m?
a. 2.0 m/s
b. 5.5 m/s
c. 7.5 m/s
d. 1.4 m/s
Answer:
d. 1.4 m/s
Explanation:
Given;
initial velocity of the `basketball, u = 4.0 m/s
acceleration of the basketball, a = -0.5 m/s² (leftward)
distance the basketball traveled, d = 14 m
the final velocity of the basketball, v = ?
(this final velocity should be less than the initial velocity since the ball is slowing down at a constant rate)
Apply the following the kinematic equation;
v² = u² + 2ad
v² = 4² + 2(-0.5)14
v² = 16 - 14
v² = 2
v = √2
v = 1.41 m/s
Therefore, the final velocity of the basketball is 1.4 m/s.
What was your train of thought as you navigated the picture of the candle?
Answer:
Where is the picture
Explanation:
WHERE IS THE PICTURE
A vehicle hits a bridge abutment at a speed estimated by
investigations as 20kmph. Skid marks of 30m on pavement
(f=0.35) followed by skid marks of 60m.on the gravel shoulder
approaching the abutment (f=0.50).What was the initial speed of vechile
Answer:
54.5 kmph
Explanation:
From work-kinetic energy principles, work done by friction on both pavement and gravel shoulder = kinetic energy change of vehicle
ΔK = W = -(f₁d₁ + f₂d₂) where f₁ = frictional force due to pavement = μ₁mg where μ₁ = coefficient of friction of pavement = 0.35, m = mass of vehicle and g = acceleration due to gravity = 9.8 m/s² and d₁ = distance moved by vehicle across pavement = 30 m and
f₂ = frictional force due to gravel shoulder = μ₂mg where μ₂ = coefficient of friction of pavement = 0.50, m = mass of vehicle and g = acceleration due to gravity = 9.8 m/s² and d₂ = distance moved by vehicle across gravel shoulder = 60 m
ΔK = 1/2m(v₁² - v₀²) where v₀ = initial velocity of vehicle, v₁ = final velocity of vehicle = 20 kmph = 20 × 1000/3600 = 5.56 m/s and m = mass of vehicle
So,
ΔK = -(f₁d₁ + f₂d₂)
1/2m(v₁² - v₀²) = -(μ₁mgd₁ + μ₂mgd₂)
1/2(v₁² - v₀²) = -(μ₁gd₁ + μ₂gd₂)
v₁² - v₀² = -2g(μ₁d₁ + μ₂d₂)
v₀² = v₁² + 2g(μ₁d₁ + μ₂d₂)
v₀ = √[v₁² + 2g(μ₁d₁ + μ₂d₂)]
substituting the values of the variables into the equation, we have
v₀ = √[(5.56 m/s)² + 2 × 9.8 m/s²(0.35 × 30 m + 0.5 × 60 m]
v₀ = √[30.91 (m/s)² + 4.9 m/s²(10.5 m + 30 m]
v₀ = √[30.91 (m/s)² + 4.9 m/s²(40.5 m]
v₀ = √[30.91 (m/s)² + 198.45 (m/s)²]
v₀ = √[229.36 (m/s)²
v₀ = 15.14 m/s
v₀ = 15.14 × 3600/1000
v₀ = 54.5 kmph
So, the initial speed of the vehicle is 54.5 kmph
PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
The first person with the right answer gets to be a brainlest
In the attachment there is a density column where there is colour
Question: tell me why is the red at the bottom of the density column if it is the least dense
Answer:
As you know, the denser objects have more weight per unit of volume, this will mean that the force that pulls down these objects is a bit larger.
This will mean that the denser objects will always go to the bottom.
This clearly implies that the red liquid, the one with one of the smaller densities, can not be at the bottom.
There are some cases where a liquid with a small density may become a lot denser as the temperature or pressure changes, and in a case like that, we could see the red liquid at the bottom, but for this case, there is no mention of changes in the temperature nor in the pressure, so this can be discarded.
The only thing that makes sense is that the red part at the bottom is the base of the tube, and has nothing to do with the red liquid.
How does temperature rise and impurities affect the surface tension of water
(2 mks)
Answer:
Surface tension is the downward force acting on the surface of liquid due to presence of inter molecular forces or cohesive forces between the particles of liquid.
Surface tension decreases with increase in temperature as the forces among particles decrease due to increase in kinetic energy and thus the cohesive nature decreases and thus surface tension also decreases.
Surface tension may decrease or increase with increase in soluble impurities .Insoluble impurities decrease the surface tension.
Determine the displacement and distance covered by a man if he walks 10 m north, turns east and walks 20 m, and then turns right and walks 10 m.
Answer:
20m
Explanation:
The two tens cancel each other out, as they are in opposite directions. Now we only care about the 20m, which if we have no 10's, will end up 20m away.
The distance covered by man is 40 m and the displacement is 20 m towards the east.
Distance and DisplacementThe distance can be defined as the measurement of the total ground covered by an object during its motion. It is a scalar quantity.
The displacement can be defined as the measurement of the shortest path over the ground is covered by an object during its motion. It is a vector quantity.
Given that a man starts walking towards the north and covered 10 m. he turns east and walks for 20 m then turns right and walks for 10 m. The attachment shows the total ground area covered by the man.
The total distance covered by the man is given below.
[tex]D = 10 + 20 + 10[/tex]
[tex]D = 40 \;\rm m[/tex]
The displacement of the man is the difference between its final position and initial position.
[tex]d = 20 - 0[/tex]
[tex]d = 20 \;\rm m[/tex]
Hence we can conclude that the distance covered by man is 40 m and the displacement is 20 m towards the east.
To know more about distance and displacement, follow the link given below.
define alpha and beta
alpha is the excess return on an investment after adjusting for market related volatility and random fluctuations.
beta is a measure of volatility relative to a benchmark ,such as the S&P 500.
Explanation:
alpha and beta are two different parts of an equation used to explain the performance of stocks and investments funds. But in maths alpha and beta is the Greek alphabet
Suppose you are on a cart that is moving at a constant speed v toward the left on a frictionless track. If you throw a massive ball straight up (from your perspective), how will the speed of the cart change?
a. The speed of the cart will increase
b. The speed of the cart will decrease
c. The speed of the cart will not change
d. You need to know how fast the ball was thrown
A 50.0-kg child stands at the rim of a merry-go-round of radius 2.25 m, rotating with an angular speed of 3.30 rad/s.. What is the child's centripetal acceleration?
Answer: the child's centripetal acceleration=24.50 m/s²
Explanation:
Given that mass of child= 50 kg
radius of merry go round= 2.25m
angular speed = 3.30 rad/s
Centripetal Acceleration = v²/ r
But V= ωr
So Centripetal Acceleration = v²/ r = (ωr)²/ r
=(3.30)² x (2.25)²/ 2.25 = (3.30)² x 2.25
=24.5025m/s²
=24.50 m/s²
Vary the sled’s height and mass. Observe the effect of each change on the potential energy of the sled.
a. How does potential energy change when height is increased?
b. How does potential energy change when mass is increased?
c. Compare a sled’s potential energy at 10 m to its potential energy at 20 m. How does doubling height affect potential energy?
d. Compare the potential energy of a 100-kg sled and a 200-kg sled at the same height. How does doubling mass affect potential energy?
Answer:
a. Potential energy of the sled is increased when height of sled is increased.
b. Potential energy of the sled is increased when height of sled is increased.
c. P.E₂₀ = 2 P.E₁₀
d. P.E₂₀₀ = 2 P.E₁₀₀
Explanation:
The potential energy of the sled can be given by the following:
[tex]Potential\ Energy = P.E = mgh\\[/tex]
where,
m = mass of sled
g = acceleration due to gravity
h = height of sled
a.
It is clear from the formula that potential energy of sled is directly proportional to the height of sled.
Therefore, potential energy of the sled is increased when height of sled is increased.
b.
It is clear from the formula that potential energy of sled is directly proportional to the mass of sled.
Therefore, potential energy of the sled is increased when mass of sled is increased.
c.
[tex]P.E\ at\ 10\ m:\\P.E_{10} = 10mg\\P.E\ at\ 20\ m:\\P.E_{20} = 20mg\\\frac{P.E_{20}}{P.E_{10}} = \frac{20mg}{10mg}[/tex]
P.E₂₀ = 2 P.E₁₀
d.
[tex]P.E\ at\ 100\ kg:\\P.E_{100} = 100gh\\P.E\ at\ 200\ m:\\P.E_{200} = 200gh\\\frac{P.E_{200}}{P.E_{100}} = \frac{200gh}{100gh}[/tex]
P.E₂₀₀ = 2 P.E₁₀₀
If -->B is added to -->A , under what conditions does the resultant vector have a magnitude equal to A + B? Under what conditions is the resultant vector equal to zero?
Answer:
See explanation below
Explanation:
For two vectors A and B to have a positive resultant, they must both move in the positive direction i.e along the positive x axis
For two vectors A and B to have a zero resultant, they must both have the same magnitude but move in the opposite direction i.e one in the positive direction while the other in the negative direction.
Fie example if A = +5N, B = -5B
So that A+B = +5-5 = 0N
You are driving at 25 m/s when an ambulance passes you and pulls into your lane, just in front of you, and speeds ahead at 35 m/s. The ambulance driver hears a siren sound of 850 Hz. What frequency do you hear
Answer:
The value is [tex]f_1 = 828 \ Hz[/tex]
Explanation:
From the question we are told that
The speed is [tex]v = 25 \ m/s[/tex]
The speed of the ambulance is [tex]u = 35 \ m/ s[/tex]
The frequency of the siren is [tex]f = 850 \ Hz[/tex]
Generally from Doppler effect equation we have that
[tex]f_1 = \frac{v_s - v }{ v_s + u } * f[/tex]
Here [tex]v_s[/tex] is the velocity of sound with the value [tex]v_s = 343 \ m/s[/tex]
=> [tex]f_1 = \frac{343 - 25 }{ 343 + 35 } * 850[/tex]
=> [tex]f_1 = 828 \ Hz[/tex]
Figure shows four paths along which objects move from a starting point to a final point, all im the same time interval. The paths pass over a grid of equally spaced straight lines. Rank the paths according to the average velocity of the objects.
Answer:
12345678901234567890
What happens when a population exceeds its carrying capacity?
Answer:
If a population exceeds carrying capacity, the ecosystem may become unsuitable for the species to survive. If the population exceeds the carrying capacity for a long period of time, resources may be completely depleted.
Show that a 2,500,000-J change in kinetic energy occurs for an airplane that is moved 500 m in takeoff by a sustained force of 5000 N.
Answer:
The answer to your question is given below
Explanation:
To solve this problem, we'll assume that the plane is initially at rest.
Hence, the kinetic energy of the plane at rest is zero i.e Initial kinetic energy (KE₁) = 0
Next, we shall determine the final kinetic energy of the plan when the force was applied. This can be obtained as follow:
Force (F) = 5000 N
Distance (s) = 500 m
Energy (E) =?
E = F × s
E = 5000 × 500
E = 2500000 J
Since energy an kinetic energy has the same unit of measurement, thus, the final kinetic energy (KE₂) of the plane is 2500000 J
Finally, we shall determine the change in the kinetic energy of the plane. This can be obtained as follow:
Initial kinetic energy (KE₁) = 0
Final kinetic energy (KE₂) = 2500000 J
Change in kinetic energy (ΔKE) =?
ΔKE = KE₂ – KE₁
ΔKE = 2500000 – 0
ΔKE = 2500000 J
Hence, the change in the kinetic energy of the plane is 2500000 J.
If a battery of 9 volts is connected across a resistor of 1000 ohm, what will be the value of current flowing through it?
A battery of 9 volts is connected to a resistor of 1000 ohms, then the value of current flowing through it will be 0.009 A.
What is current?Any flow of electrical charge carriers, such as ions, holes, or subatomic charged particles, is referred to as electrical current.
When electrons serve as the charge transfer in a wire, the amount of charge moving through any point of wire in a given amount of time is measured as electric current. Electric charges' motion is intermittently reversed in the alternating current, but not in direct current.
In many situations, the direction of movement in electric circuits is assumed to be the direction of positive charge flow, which is the way opposite of the real particle drift. When properly specified, the current is known as conventional current.
From the question,
V = ir
i = v/r
i = 9/1000
i = 0.009 A
Therefore, the value of the current flowing through it is 0.009 A.
To know more about Current :
https://brainly.com/question/13076734
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a planet has been detected in a circular orbit around the star Rho1 Cancri with an orbital radius equal to 1.65 x 10^10 m. the orbital period of this planet is approximately 14.5 days which is the approximate mass of the star pho1 cancri
Answer:
Approximately [tex]1.69 \times 10^{30}\; \rm kg[/tex].
Explanation:
Deduction of the formulaLet [tex]M[/tex] and [tex]m[/tex] denote the mass of the star and the planet, respectively.
Let [tex]G[/tex] denote the constant of universal gravitation ([tex]G \approx 6.67408 \times 10^{-11}\; \rm m^{3} \cdot kg^{-1}\cdot s^{-2}[/tex].)
Let [tex]r[/tex] denote the orbital radius of this planet (assuming that [tex]r\![/tex] is constant.) The question states that [tex]r = 1.65 \times 10^{10}\; \rm m[/tex].
The size of gravitational attraction of the star on this planet would be:[tex]\displaystyle \frac{G \cdot M \cdot m}{r^{2}}[/tex].
If attraction from the star is the only force on this planet, the net force on this planet would be [tex]\displaystyle \frac{G \cdot M \cdot m}{r^{2}}[/tex].
Let [tex]\omega[/tex] denote the angular velocity of this planet as it travels along its circular orbit around the star. The size of [tex]\omega\![/tex] could be found from the period [tex]T[/tex] of each orbit: [tex]\omega = (2\, \pi) / T[/tex].
In other words, this planet of mass [tex]m[/tex] is in a circular motion with radius [tex]r[/tex] and angular velocity [tex]\omega[/tex]. Therefore, the net force on this planet should be equal to [tex]m \cdot \omega^2 \cdot r[/tex].
Hence, there are two expressions for the net force on this planet:
[tex]\text{Net Force} = \displaystyle \frac{G \cdot M \cdot m}{r^{2}}[/tex] from universal gravitation, and[tex]\displaystyle \text{Net Force} = m \cdot \omega^2 \cdot r = {\left(\frac{2\pi}{T}\right)}^{2} m \cdot r[/tex] from circular motion.Equate the right-hand side of these two equations:
[tex]\displaystyle \frac{G \cdot M \cdot m}{r^2} = {\left(\frac{2\pi}{T}\right)}^{2}\, m \cdot r[/tex].
Simplify this equation and solve for [tex]M[/tex], the mass of the star:
[tex]\displaystyle M = \frac{{(2\pi / T)}^2 \cdot r^3}{G}[/tex].
Notice that [tex]m[/tex], the mass of the planet, was eliminated from the equation. That explains why this question could be solved without knowing the exact mass of the observed planet.
Actual CalculationsConvert the orbital period of this star to standard units:
[tex]\begin{aligned}T &= 14.5\; \text{day} \times \frac{24\; \text{hour}}{1\; \text{day}} \times \frac{3600\; \text{second}}{1\; \text{hour}} \\ & = 1.2528 \times 10^{6}\; \rm \text{second}\end{aligned}[/tex].
Calculate the mass of the star:
[tex]\begin{aligned} M &= \frac{{(2\pi / T)}^2 \cdot r^3}{G} \\ &\approx \frac{\displaystyle {\left(\frac{2\pi}{1.2528 \times 10^{6}\; \rm s}\right)}^{2} \times \left(1.65 \times 10^{10}\; \rm m\right)^{3}}{6.67408 \times 10^{-11}\; \rm m^{3}\cdot kg^{-1} \cdot s^{-2}}\\ &\approx 1.69 \times 10^{30}\; \rm kg\end{aligned}[/tex].
An ideal spring with spring constant k is hung from the ceiling. The initial length of the spring, with nothing attached to the free end, is 15.0 cm. When an object of mass m is attached to the free end of the spring, the spring stretches to a final length of 30.0 cm. If k = 3.5 N/cm, what is the mass m of the object?
The mass m of the object = 5.25 kg
Further explanationGiven
k = spring constant = 3.5 N/cm
Δx= 30 cm - 15 cm = 15 cm
Required
the mass m
Solution
F=m.g
Hooke's Law
F = k.Δx
[tex]\tt m.g=k.\Delta x\\\\m.10=3.5\times 15\\\\m=5.25~kg[/tex]