Answer:
The mass of the central black hole is [tex]7.19x10^{37} Kg[/tex]
Explanation:
The Universal law of gravitation shows the interaction of gravity between two bodies:
[tex]F = G\frac{Mm}{r^{2}}[/tex] (1)
Where G is the gravitational constant, M and m are the masses of the two objects and r is the distance between them.
For this particular case M is the mass of the central black hole and m is the mass of the molecular cloud. Since it is a circular motion the centripetal acceleration will be:
[tex]a = \frac{v^{2}}{r}[/tex] (2)
Then Newton's second law ([tex]F = ma[/tex]) will be replaced in equation (1):
[tex]ma = G\frac{Mm}{r^{2}}[/tex]
By replacing (2) in equation (1) it is gotten:
[tex]m\frac{v^{2}}{r} = G\frac{Mm}{r^{2}}[/tex] (3)
Therefore, the mass of the central black hole can be determined if M is isolated from equation (3):
[tex]M = \frac{rv^{2}}{G}[/tex] (4)
Equation 4 it is known as the orbital velocity law.
Where M is the mass of the central black hole, r is the orbital radius, v is the orbital speed and G is the gravitational constant.
Before replacing the orbital speed in equation 4 it is necessary to make the conversion from kilometers to meters:
[tex]1000 \frac{km}{s} x \frac{1000 m}{1 km}[/tex] ⇒ [tex]1000000 m/s[/tex]
Then, equation 4 can be finally used:
[tex]M = \frac{(4.8x10^{15} m)(1000000 m/s)^{2}}{(6.67x10^{-11} N.m^{2}/Kg^{2})}[/tex]
[tex]M = 7.19x10^{37} Kg[/tex]
Hence, the mass of the central black hole is [tex]7.19x10^{37} Kg[/tex]
To calculate the mass of the central black hole, we can use the orbital velocity law, Mr = r×v^2/G. Given the orbital radius and velocity of the molecular clouds, we can plug these values into the equation to find the mass of the black hole.
Explanation:To calculate the mass of the central black hole, we can use the orbital velocity law: Mr=r×v^2/G. In this equation, Mr represents the mass of the black hole, r is the orbital radius, v is the orbital velocity, and G is the gravitational constant. Given that the orbital radius is 0.49 light-years (4.8x10^15 meters) and the orbital velocity is 1000 km/s, we can plug these values into the equation to find the mass of the black hole.
First, let's convert the orbital velocity from km/s to m/s: 1000 km/s = 1000 x 1000 m/s = 1,000,000 m/s. Now, we plug in the values to the equation:
Mr = (4.8x10^15 meters) x (1,000,000 m/s)^2 / (6.67430 × 10^-11 m^3⋅kg^−1⋅s^−2)
Simplifying the expression, we get:
Mr = (4.8x10^15 meters) x (1,000,000 m/s)^2 / (6.67430 × 10^-11) kg
After calculating, the mass of the central black hole is approximately 3.23x10^37 kg.
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A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2 . a) After 2 seconds of acceleration, how many radians of rotation has the wheel completed? b) After 3 seconds of acceleration, what is the wheel’s kinetic energy due to rotation?
Answer:
(A) Angular speed 40 rad/sec
Rotation = 50 rad
(b) 37812.5 J
Explanation:
We have given moment of inertia of the wheel [tex]I=25kgm^2[/tex]
Initial angular velocity of the wheel [tex]\omega _0=10rad/sec[/tex]
Angular acceleration [tex]\alpha =15rad/sec^2[/tex]
(a) We know that [tex]\omega =\omega _0+\alpha t[/tex]
We have given t = 2 sec
So [tex]\omega =10+15\times 2=40rad/sec[/tex]
Now [tex]\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad[/tex]
(b) After 3 sec [tex]\omega =10+15\times 3=55rad/sec[/tex]
We know that kinetic energy is given by [tex]Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J[/tex]
Two children are balanced on opposite sides of a seesaw. If one child leans inward toward the pivot point, her side will
a. fall.
b. rise.
c. neither rise nor fall.
From the definition of equilibrium, at the moment in which both children are sitting facing each other at a certain distance the torque within the seesaw will be zero.
However, if one of the children approaches the pivot, the center of mass will move towards the end of the other child, which will immediately cause the child to rise.
Another way of observing this problem is considering the distance between the two, the distance is proportional to the Torque,
[tex]\tau_1 = \tau_2[/tex]
[tex]F*d_1 = F*d_2[/tex]
Therefore by decreasing the distance - when walking towards the pivot - the torque of the child sitting at the other end will be greater because it keeps its distance.
Being said higher torque will cause the approaching child to rise.
The correct answer is B.
The child's side will rise because of shifting of weight of the child from its place.
How the child's side rise?Two children are balanced on opposite sides of a seesaw. If one child leans inward toward the pivot point, her side will rise because of shifting of weight of the child from its place. Due to movement from its place, the weight of other child increases which leads to lowering of its side.
So we can conclude that the child's side will rise because of shifting of weight of the child from its place.
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The matter that makes up a planet is distributed uniformly so that the planet has a fixed, uniform density. How does the magnitude of the acceleration due to gravity g at the planet surface depend on the planet radius R ? (Hint: how does the total mass scale with radius?) g ∝ 1 / R g ∝ R g ∝ √ R g ∝ R 2
Answer:
the acceleration due to gravity g at the surface is proportional to the planet radius R (g ∝ R)
Explanation:
according to newton's law of universal gravitation ( we will neglect relativistic effects)
F= G*m*M/d² , G= constant , M= planet mass , m= mass of an object , d=distance between the object and the centre of mass of the planet
if we assume that the planet has a spherical shape, the object mass at the surface is at a distance d=R (radius) from the centre of mass and the planet volume is V=4/3πR³ ,
since M= ρ* V = ρ* 4/3πR³ , ρ= density
F = G*m*M/R² = G*m*ρ* 4/3πR³/R²= G*ρ* 4/3πR
from Newton's second law
F= m*g = G*ρ*m* 4/3πR
thus
g = G*ρ* 4/3π*R = (4/3π*G*ρ)*R
g ∝ R
A runner of mass 56.0kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude 3.10m/s . The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.210rad/s relative to the earth. The radius of the turntable is 3.10m , and its moment of inertia about the axis of rotation is 81.0kg?m2
Question: Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can treat the runner as a particle.)
This problem can be resolved using the principle of conservation of angular momentum. Initially, the positive momentum of runner and negative of turntable cancel each other, making the total momentum zero. When the runner stops, the angular momentum of the turntable is expected to compensate for the total momentum of the system.
Explanation:In Physics, this is a classic problem involving the conservation of angular momentum. Angular momentum is given as L = Iω for an object rotating about an axis, where I is the moment of inertia and ω is the angular velocity. The total initial angular momentum of the system is zero because the positive runner's angular momentum cancels out the negative turntable's angular momentum.
When the runner comes to rest, he no longer contributes an angular momentum. The turntable has to account for all of the angular momentum. We therefore use the conservation of angular momentum to solve for the new angular momentum of the turntable after the runner stops:
Equating moment of inertia of the runner (considering the runner as a point mass, moment of inertia, I = m*r^2) and moment of inertia of the turntable and using the equation L_initial = L_final, we can find the final angular velocity of the system:
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The final angular velocity of the system is 6.853 rad/s.
The initial angular momentum [tex]\(L_{i}\)[/tex] of the turntable is:
[tex]\[ L_{i} = I\omega_{i} \][/tex]
where I is the moment of inertia of the turntable.
The initial angular momentum [tex]\(L_{runner,i}\)[/tex] of the runner is:
[tex]\[ L_{runner,i} = mvr \][/tex]
since the angular momentum of a particle moving in a circle is the product of its mass, velocity, and the radius of the circle.
The final angular momentum [tex]\(L_{f}\)[/tex] of the system is:
[tex]\[ L_{f} = I\omega_{f} + mvr_{f} \][/tex]
where [tex]\(r_{f}\)[/tex] is the final radius at which the runner is at rest relative to the turntable, which is the same as the initial radius r since the runner is still on the turntable.
By conservation of angular momentum:
[tex]\[ L_{i} + L_{runner,i} = L_{f} \] \[ I\omega_{i} + mvr = I\omega_{f} + mvr \][/tex]
Since the runner is at rest relative to the turntable in the final state, (vr = 0), and the equation simplifies to:
[tex]\[ I\omega_{i} + mvr = I\omega_{f} \][/tex]
Now we can solve for [tex]\(\omega_{f}\)[/tex]:
[tex]\[ \omega_{f} = \frac{I\omega_{i} + mvr}{I} \] \[ \omega_{f} = \omega_{i} + \frac{mvr}{I} \][/tex]
Given:
[tex]\(m = 56.0\) kg\\ \(v = 3.10\) m/s\\ \(r = 3.10\) m\\ \(I = 81.0\) kg\(\cdot\)m\(^2\)\\ \(\omega_{i} = 0.210\) rad/s[/tex]
Plugging in the values:
[tex]\[ \omega_{f} = 0.210 + \frac{56.0 \times 3.10 \times 3.10}{81.0} \] \[ \omega_{f} = 0.210 + \frac{56.0 \times 9.61}{81.0} \] \[ \omega_{f} = 0.210 + \frac{538.16}{81.0} \] \[ \omega_{f} = 0.210 + 6.643 \] \[ \omega_{f} = 6.853 \][/tex]
An object moves in simple harmonic motion with amplitude 14cm and period 0.2 seconds. At time =t0 seconds, its displacement d from rest is 0cm, and initially it moves in a positive direction. Give the equation modeling the displacement d as a function of time t.
Answer:
d= 14 sin 31.41 t
Explanation:
Lets take the general equation of the motion
d= D sinωt
d=Displacement
ω=Natural angular frequency
t=Time
D=Amplitude
Given that D= 14 cm
We know that time period T
[tex]T=\dfrac{2\pi}{\omega}[/tex]
Given that T= 0.2 s
[tex]0.2=\dfrac{2\pi}{\omega}[/tex]
ω=31.41 rad/s
D= 14 cm
Therefore
d= D sinωt
d= 14 sin 31.41 t
Final answer:
The equation modeling the displacement d of an object in simple harmonic motion with an amplitude of 14 cm and a period of 0.2 seconds, starting at 0 cm and moving in a positive direction at t = 0, is d(t) = 14cos(10πt + π/2) cm.
Explanation:
The displacement d of an object in simple harmonic motion can be modeled by the function d(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase constant. Since the amplitude A is 14 cm and the period T is 0.2 seconds, and given that the initial displacement at time t = 0 is 0 cm and it initially moves in a positive direction, we can determine ω using the formula ω = 2π/T and φ using the fact that the cosine function must start at 0.
Therefore, the angular frequency ω = 2π/0.2 s = 10π rad/s, and since the function starts at 0 and moves in a positive direction, φ = +π/2. Hence, the equation that models the displacement d as a function of time t is d(t) = 14cos(10πt + π/2) cm.
A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an additional 3.0 cm downward and released, what is the period of the resulting oscillation? Give your answer in seconds with 3 significant figures. A = 13 B = 427
Answer:
Time period of the osculation will be 2.1371 sec
Explanation:
We have given mass m = (B+25)
And the spring is stretched by (8.5 A )
Here A = 13 and B = 427
So mass m = 427+25 = 452 gram = 0.452 kg
Spring stretched x= 8.5×13 = 110.5 cm
As there is additional streching of spring by 3 cm
So new x = 110.5+3 = 113.5 = 1.135 m
Now we know that force is given by F = mg
And we also know that F = Kx
So [tex]mg=Kx[/tex]
[tex]K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m[/tex]
Now we know that [tex]\omega =\sqrt{\frac{K}{m}}[/tex]
So [tex]\frac{2\pi }{T} =\sqrt{\frac{K}{m}}[/tex]
[tex]\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}[/tex]
[tex]T=2.1371sec[/tex]
The period of the resulting oscillation, named T, given a mass of (B + 25.0) g (with B = 427 g) and a spring displacement of (8.50 A) cm (with A = 13), when the object is pulled an additional 3.0 cm downward and released, is 2.345 seconds, when rounded to three significant figures.
Explanation:The problem given can be solved using the concept of simple harmonic motion, which is encountered in Physics. The object attached to the spring can be visualized as a simple harmonic oscillator. The period of the resulting oscillation can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass and k is the spring constant.
First, we compute the mass in grams, which is given by (B + 25.0) g, where B = 427 g. Therefore, the mass is (427 g + 25.0 g) = 452 g or 0.452 kg (since 1 kg = 1000 g).
Next, the spring constant k is calculated using Hooke’s law (F=kx), where F is the force exerted by the mass on the spring (F=mg), m is the mass and g is the acceleration due to gravity, and x is the displacement of the spring. From the problem, m = 0.452 kg, g = 9.8 m/s², and x = (8.50 A) cm = (8.50 x 13) cm = 110.5 cm or 1.105 m (since 1 m = 100 cm). Substituting these values into Hooke’s law gives: k = mg/x = (0.452 kg x 9.8 m/s²) / 1.105 m = 4.00 N/m.
Finally, the period T of the resulting oscillation is found using the formula T = 2π√(m/k) = 2π√(0.452 kg/4.00 N/m) = 2.345 s, rounded to three significant figures.
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Which of the following is prohibited by the 2nd law of thermodynamics?
(A) A device that converts heat into work with 100% efficiency
(B) A device that converts work into heat with 100% efficiency.
(C) A device that uses work to supply X heat at high temperature while taking in Y heat at lower temperature. where X>Y.
(D) A device that converts work into heat with >100% efficiency
(A) A device that converts heat into work with 100% efficiency
It clearly violates the second law of thermodynamics because it warns that while all work can be turned into heat, not all heat can be turned into work. Therefore, despite the innumerable efforts, the efficiencies of the bodies have only been able to reach 60% at present.
A 0.32 kg turntable of radius 0.18 m spins about a vertical axis through its center. A constant rotational acceleration causes the turntable to accelerate from 0 to 24 revolutions per second in 8.0 s. [Hint: model the turntable has a uniform density disk.] What is the rotational acceleration of the turntable
Answer:[tex]18.85 rad/s^2[/tex]
Explanation:
Given
mass of turntable [tex]m=0.32 kg[/tex]
radius [tex]r=0.18 m[/tex]
[tex]N=24 rev/s[/tex]
[tex]\omega =2\pi N=2\pi 24[/tex]
[tex]\omega =48\pi rad/s[/tex]
time interval [tex]t=8 s[/tex]
using [tex]\omega =\omeag _0+\alpha t[/tex]
where [tex]\omega =final\ angular\ velocity[/tex]
[tex]\omega _0=initial\ angular\ velocity[/tex]
[tex]\alpha =angular\ acceleration[/tex]
[tex]t=time[/tex]
[tex]48\pi =0+\alpha \times 8[/tex]
[tex]\alpha =6\pi rad/s^2[/tex]
[tex]\alpha =18.85 rad/s^2[/tex]
How does the magnitude of the electrical force compare between a pair of charged particles when they are brought to half their original distance of separation? To one-quarter their original distance? To four times their original distance? (What law guides your answers?)
Answer:
a) 4 times larger. b) 16 times larger. c) 16 times smaller. d) Coulomb´s Law
Explanation:
Between any pair of charged particle, there exists a force, acting on the line that join the charges (assuming they can assimilated to point charges) directed from one to the other, which is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them.
F= k q1q2 / (r12)2
a) If the distance is reduced to the half of the original distance of separation, and we introduce this value in the force equation, we get:
F(r/2) = k q1q2 / (r12/2)2 = k q1q2 /((r12)2/4) = 4 F(r)
b) By the same token, if r= r/4, we will have F(r/4) = 16 F(r)
c) If the distance increases 4 times, as the force is inversely proportional to the square of the distance, the force will be the original divided by 16, i.e., 16 times smaller.
The empirical law that allows to find out easily these values, is the Coulomb´s Law.
Explanation:
5. A 55-kg swimmer is standing on a stationary 210-kg floating raft. The swimmer then runs off the raft horizontally with the velocity of +4.6 m/s relative to the shore. Find the recoil velocity that the raft would have if there were no friction and resistance due to the water.
Answer:
The recoil velocity of the raft is 1.205 m/s.
Explanation:
given that,
Mass of the swimmer, [tex]m_1=55\ kg[/tex]
Mass of the raft, [tex]m_2=210\ kg[/tex]
Velocity of the swimmer, v = +4.6 m/s
It is mentioned that the swimmer then runs off the raft, the total linear momentum of the swimmer/raft system is conserved. Let V is the recoil velocity of the raft.
[tex]m_1v+m_2V=0[/tex]
[tex]55\times 4.6+210V=0[/tex]
V = -1.205 m/s
So, the recoil velocity of the raft is 1.205 m/s. Hence, this is the required solution.
Answer:
The recoil velocity of the raft would be [tex]v_{r}\approx 1.2\frac{m}{s}[/tex] (pointing to the left if the swimmer runs to the right)
Explanation:
The problem states that the swimmer has a mass of m=55 kg, and the raft has a mass of M=210 kg. Then, it says that the swimmer runs off the raft with a (final) velocity of v=4.6 m/s relative to the shore.
To analyze it, we take a system of "two particles", wich means that we will consider the swimmer and the raft as a hole system, aisolated from the rest of the world.
Then, from the shore, we can put our reference system and take the initial moment when the swimmer and the raft are stationary. This means that the initial momentum is equal to zero:
[tex]p_{i}=0[/tex]
Besides, we can use magnitudes instead of vectors because the problem will develope in only one dimension after the initial stationary moment (x direction, positive to the side of the running swimmer, and negative to the side of the recoling raft), this means that we can write the final momentum as
[tex]p_{f}=mv-Mv_{r}=0[/tex]
The final momentum is equal to zero due to conservation of momentum (because there are no external forces in the problem, for the system "swimmer-raft"), so the momentum is constant.
Then, from that previous relation we can clear
[tex]v_{r}=\frac{m}{M}v=\frac{55}{210}*4.6\frac{m}{s}=\frac{253}{210}\frac{m}{s}\approx1.2\frac{m}{s}[/tex]
wich is the recoil velocity of the raft, and it is pointing to the left (we established this when we said that the raft was going to the negative side of the system of reference, and when we put a minus in the raft term inside the momentum equation).
Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignoring Earth's rotation, how much energy is needed to get the same object into orbit at height h?Express your answer in terms of some or all of the variables h, mass of the object m, mass of Earth mE, its radius RE, and gravitational constant G.
Answer:
Explanation:
Potential energy on the surface of the earth
= - GMm/ R
Potential at height h
= - GMm/ (R+h)
Potential difference
= GMm/ R - GMm/ (R+h)
= GMm ( 1/R - 1/ R+h )
= GMmh / R (R +h)
This will be the energy needed to launch an object from the surface of Earth to a height h above the surface.
Extra energy is needed to get the same object into orbit at height h
= Kinetic energy of the orbiting object at height h
= 1/2 x potential energy at height h
= 1/2 x GMm / ( R + h)
Tarzan is running with a horizontal velocity, along level ground. While running, he encounters a 2.21 m vine of negligible mass, the vine was hanging vertically from a tall tree limb and the end of the vine is directly level with his center of mass. Tarzan decides to grab the vine and swing which causes him to experience uniform circular motion. At the maximum height, Tarzan notices that the angle the vine makes with the vertical is 19 degrees. Tarzan mass is 65kg. Ignoring any elastic properties of the vine may have.a) Determine the magnitude of Tarzan's centripetal accelerationb) Determine the magnitude of the tension at the beginning of the swing
Answer:
a) [tex]a_c = 1.09m/s^2[/tex]
b) T = 720.85N
Explanation:
With a balance of energy from the lowest point to its maximum height:
[tex]m*g*L(1-cos\theta)-1/2*m*V_o^2=0[/tex]
Solving for [tex]V_o^2[/tex]:
[tex]V_o^2=2*g*L*(1-cos\theta)[/tex]
[tex]V_o^2=2.408[/tex]
Centripetal acceleration is:
[tex]a_c = V_o^2/L[/tex]
[tex]a_c = 2.408/2.21[/tex]
[tex]a_c = 1.09m/s^2[/tex]
To calculate the tension of the rope, we make a sum of forces:
[tex]T - m*g = m*a_c[/tex]
Solving for T:
[tex]T =m*(g+a_c)[/tex]
T = 720.85N
The heart produces a weak magnetic field that can be used to diagnose certain heart problems. It is a dipole field produced by a current loop in the outer layers of the heart.
a. It is estimated that the field at the center of the heart is 90 pT. What current must circulate around an 8.0-cm-diameter loop, about the size of a human heart, to produce this field?
To produce a magnetic field of 90 pT at the center of an 8.0-cm-diameter loop, a current of approximately 0.045 Amps must circulate around the loop.
Explanation:To calculate the current needed to produce a magnetic field of 90 pT at the center of an 8.0-cm-diameter loop, we can use the formula for the magnetic field strength at the center of a circular loop. The formula is given by B = (µ₀I)/(2R), where B is the magnetic field strength, µ₀ is the permeability of free space, I is the current, and R is the radius of the loop.
First, let's convert 90 pT to Tesla by dividing by 10^12. This gives us a value of 9 x 10^-11 T. We can now substitute this value into the formula along with the given radius of 4.0 cm (half of the diameter) to solve for the current.
B = (µ₀I)/(2R)
9 x 10^-11 T = (4π x 10^-7 T m/A)(I)/(2(0.04 m))
Simplifying the equation, we find:
I = (2π(0.04 m)(9 x 10^-11 T))/(4π x 10^-7 T m/A) = 0.045 A
Therefore, a current of approximately 0.045 Amps must circulate around the 8.0-cm-diameter loop to produce a magnetic field of 90 pT at its center.
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If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350 m from the person's eyes? Express your answer in meters.
Answer:
f = 0.84 m
Explanation:
given,
near point of the distance from eye(di) = 0.60 m
focal length calculation = ?
distance he can read (d_o)= 0.350 m
now,
Using lens formula
[tex]\dfrac{1}{f} = \dfrac{1}{d_0} + \dfrac{1}{d_i}[/tex]
[tex]\dfrac{1}{f} = \dfrac{1}{0.35} + \dfrac{1}{-0.6}[/tex]
[tex]\dfrac{1}{f} = 1.19[/tex]
[tex] f = \dfrac{1}{1.19}[/tex]
f = 0.84 m
The focal length of the lens f = 0.84 m
In a series RCL circuit the generator is set to a frequency that is not the resonant frequency.
This nonresonant frequency is such that the ratio of the inductive reactance to the capacitive reactance of the circuit is observed to be 6.72. The resonant frequency is 240 Hz.
What is the frequency of the generator?
Answer: 624 Hz
Explanation:
If the ratio of the inductive reactance to the capacitive reactance, is 6.72, this means that it must be satified the following expression:
ωL / 1/ωC = 6.72
ω2 LC = 6.72 (1)
Now, at resonance, the inductive reactance and the capacitive reactance are equal each other in magnitude, as follows:
ωo L = 1/ωoC → ωo2 = 1/LC
So, as we know the resonance frequency, we can replace LC in (1) as follows:
ω2 / ωo2 = 6.72
Converting the angular frequencies to frequencies, we have:
4π2 f2 / 4π2 fo2 = 6.72
Simplifying and solving for f, we have:
f = 240 Hz . √6.72 = 624 Hz
As the circuit is inductive, f must be larger than the resonance frequency.
A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. Even without knowing how big the pressure cooker is, or what altitude it is being used at, we can make predictions about how much force the lid will experience under different conditions. Part A If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the magnitude of the net force F120 on the lid when the air inside the cooker had been heated to 120∘C? Assume that the temperature of the air outside the pressure cooker is 20∘C (room temperature) and that the area of the pressure cooker lid is A. Take atmospheric pressure to be pa. Treat the air, both inside and outside the pressure cooker, as an ideal gas obeying pV=NkBT. Express the force in terms of given variables. View Available Hint(s) F120F 120 F_120 = nothing Part B The pressure relief valve on the lid is now opened, allowing hot air to escape until the pressure inside the cooker becomes equal to the outside pressure pa. The pot is then sealed again and removed from the stove. Assume that when the cooker is removed from the stove, the air inside it is still at 120∘C. What is the magnitude of the net force F20 on the lid when the air inside the cooker has cooled back down to 20∘C?
The net force on the pressure cooker lid depends on the temperature inside the cooker. At 120°C, the net force is found using the ideal gas law and the given parameters. After the cooker cools to 20°C, and pressure is equalized, the net force is zero.
Explanation:Part A: The net force on the pressure cooker lid at 120°C is dependent on the difference in pressure inside and outside the cooker. Using the ideal gas law, we find pressure is proportional to temperature (p = NkBT/V). The force exerted on the cooker lid is the product of this pressure and the lid's area (F = pA). Given that the pressure outside the pot is pa (atmospheric pressure) at a temperature of 20°C and the pressure inside increased to a temperature of 120°C, the net force on the lid is F120 = A(NkB(120+273)/V - pa), where T is in Kelvins.
Part B: After the pressure equalizes and the cooker has cooled down to 20°C, the pressures inside and outside the cooker are equal, therefore the net force is zero. This is because the difference in pressures is what causes the force (F20 = A(pa - pa) = 0)
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In large buildings, hot water in a water tank is circulated through a loop so that the user doesn’t have to wait for all the water in long piping to drain before hot water starts coming out. A certain recirculating loop involves 40-m-long, 1.2-cm-diameter cast iron pipes with six 90° threaded smooth bends and two fully open gate valves. If the average flow velocity through the loop is 2 m/s, determine the required power input for the recirculating pump. Take the average water temperature to be 60°C and the efficiency of the pump to be 76 percent. The density and dynamic viscosity of water at 60°C are rho = 983.3 kg/m3, μ = 0.467 × 10–3 kg/m·s. The roughness of cast iron pipes is 0.00026 m. The loss coefficient is KL = 0.9 for a threaded 90° smooth bend and KL = 0.2 for a fully open gate valve. (Round the final answer to three decimal paces.)
Answer:
The power input is 0.102 kW
Solution:
As per the question:
Length of the loop, L = 40 m
Diameter of the loop, d = 1.2 cm
Velocity, v = 2 m/s
Loss coefficient of the threaded bends, [tex]K_{L, bend} = 0.9[/tex]
Loss coefficient of the valve, [tex]K_{L, valve} = 0.2[/tex]
Dynamic viscosity of water, [tex]\mu = 0.467\times 10^{- 3}\ kg/m.s[/tex]
Density of water, [tex]\rho = 983.3\ kg/m^{3}[/tex]
Roughness of the pipe of cast iron, [tex]\epsilon = 0.00026\ m[/tex]
Efficiency of the pump, [tex]\eta = 0.76[/tex]
Now,
We calculate the volume flow rate as:
[tex]\dot{V} = Av[/tex]
where
[tex]\dot{V}[/tex] = Volume rate flow
A = Area
v = velocity
[tex]\dot{V} = \frac{\pi}{4}d^{2}\times 2 = 2.262\times 10^{- 4}\ m^{3}/s[/tex]
For this, Reynold's N. is given by:
[tex]R_{e} = \frac{\rho vd}{\mu}[/tex]
[tex]R_{e} = \frac{983.3\times 2\times 0.012}{0.467\times 10^{- 3}} = 50533.62[/tex]
Since, [tex]R_{e}[/tex] > 4000, the flow is turbulent in nature.
Now,
With the help of the Colebrook eqn, we calculate the friction factor as:
[tex]\frac{1}{\sqrt{f}} = - 2log[\frac{\frac{\epsilon}{d}}{3.7} + \frac{2.51}{R_{e}\sqrt{f}}][/tex]
[tex]\frac{1}{\sqrt{f}} = - 2log[\frac{\frac{0.00026}{0.012}}{3.7} + \frac{2.51}{50533.62\sqrt{f}}][/tex]
f = 0.05075
Now,
To calculate the total head loss:
[tex]H_{loss} = (\frac{fL}{d} + 6K_{L, bend} + 2K_{L, valve})\farc{v^{2}}{2g}[/tex]
[tex]H_{loss} = (\frac{0.05075\times 40}{0.012} + 6\times 0.9 + 2\times 0.2)\farc{2^{2}}{2\times 9.8} = 35.71\ m[/tex]
Now,
The drop in the pressure can be calculated as:
[tex]\Delat P = \rho g H_{loss}[/tex]
[tex]\Delat P = 983.3\times 9.8\times 35.71 = 344.113\ kN/m^{2}[/tex]
Now,
to calculate the input power:
[tex]\dot{W} = \frac{\dot{W_{p}}}{\eta}[/tex]
[tex]\dot{W} = \frac{\dot{V}\Delta P}{0.76}[/tex]
[tex]\dot{W} = \frac{2.262\times 10^{- 4}\times 344.113\times 1000}{0.76} = 0.102\ kW[/tex]
A continuous succession of sinusoidal wave pulses are produced at one end of a very long string and travel along the length of the string. The wave has frequency 62.0 Hz, amplitude 5.20 mm and wavelength 0.560 m.
(a) How long does it take the wave to travel a distance of 8.40 m along the length of the string?
(b) How long does it take a point on the string to travel a distance of 8.40 m, once the wave train has reached the point and set it into motion?
(c) In parts (a) and (b), how does the time change if the amplitude is doubled?
a) The time taken for the wave to travel a distance of 8.40 m is 0.24 s
b) The time taken for a point on the string to travel a distance of 8.40 m is 6.46 s
c) The time in part a) does not change; the time in part b) is halved (3.23 s)
Explanation:
a)
In order to solve this part, we have to find the speed of the wave, which is given by the wave equation:
[tex]v=f \lambda[/tex]
where
v is the speed
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the wave in this problem,
[tex]f = 62.0 Hz\\\lambda =0.560 m[/tex]
So its speed is
[tex]v=(62.0)(0.560)=34.7 m/s[/tex]
Now we want to know how long does it take for the wave to travel a distance of
d = 8.40 m
Since the wave has a constant velocity, we can use the equation for uniform motion:
[tex]t=\frac{d}{v}=\frac{8.40}{34.7}=0.24 s[/tex]
b)
In this case, we want to know how long does it take for a point on the string to travel a distance of
d = 8.40 m
A point on the string does not travel along the string, but instead oscillates up and down. The amplitude of the wave corresponds to the maximum displacement of a point on the string from the equilibrium position, and for this wave it is
A = 5.20 mm = 0.0052 m
The period of the wave is the time taken for a point on the string to complete one oscillation. It can be calculated as the reciprocal of the frequency:
[tex]T=\frac{1}{f}=\frac{1}{62.0}=0.016 s[/tex]
During one oscillation (so, in a time corresponding to one period), the distance covered by a point on the string is 4 times the amplitude:
[tex]d=4A=4(0.0052)=0.0208 m[/tex]
Since d is the distance covered by a point on the string in a time T, then we can find the time t needed to cover a distance of
d' = 8.40 m
By setting up the following proportion:
[tex]\frac{d}{T}=\frac{d'}{t}\\t=\frac{d'T}{d}=\frac{(8.40)(0.016)}{0.0208}=6.46 s[/tex]
c)
As we can see from the equation used in part a), the speed of the wave depends only on its frequency and wavelength, not on the amplitude: therefore, if the amplitude is changed, the speed of the wave is not affected, therefore the time calculated in part a) remains the same.
On the contrary, the answer to part b) is affected by the amplitude. In fact, if the amplitude is doubled:
A' = 2A = 2(0.0052) = 0.0104 m
Then the distance covered during one period is
[tex]d=4A' = 4(0.0104)=0.0416 m[/tex]
The time period does not change, so it is still
T = 0.016 s
Therefore, the time needed to cover a distance of
d' = 8.40 m
this time is
[tex]\frac{d}{T}=\frac{d'}{t}\\t=\frac{d'T}{d}=\frac{(8.40)(0.016)}{0.0416}=3.23 s[/tex]
So, the time taken has halved.
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The time taken for a wave to travel a distance of 8.40 m is approximately 0.120 seconds. The time taken for a point on the string to travel 8.40 m once the wave train has reached it is also approximately 0.120 seconds. Doubling the amplitude of the wave does not affect the time taken.
Explanation:To determine how long it takes for the wave to travel a distance of 8.40 m, we can use the formula:
Time = Distance / Speed
The speed of a wave is equal to the product of its frequency and wavelength:
Speed = Frequency x Wavelength
Substituting the given values into the formulas, we find that it takes approximately 0.120 seconds for the wave to travel 8.40 m along the string.
To calculate the time it takes for a point on the string to travel a distance of 8.40 m once the wave train has reached the point and set it into motion, we need to divide the distance by the speed of the wave pulse:
Time = Distance / Speed
Since the speed of the wave pulse is the same as the speed of the wave, the time it takes for a point on the string to travel 8.40 m is also approximately 0.120 seconds.
If the amplitude of the wave is doubled, it does not affect the time it takes for the wave to travel a distance of 8.40 m or for a point on the string to travel 8.40 m. The only thing that changes is the maximum displacement of the particles in the wave, but it does not impact the time taken.
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A Texas cockroach of mass 0.117 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has a radius 17.3 cm, rotational inertia 5.20 x 10-3 kg·m2, and frictionless bearings. The cockroach's speed (relative to the ground) is 1.91 m/s, and the lazy Susan turns clockwise with angular velocity ω0 = 2.87 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
Answer
given,
mass of cockroach = 0.117 Kg
radius = 17.3 cm
rotational inertia = 5.20 x 10⁻³ Kg.m²
speed of cockroach = 1.91 m/s
angular velocity of Susan (ω₀)= 2.87 rad/s
final speed of cockroach = 0 m/s
Initial angular velocity of Susan
L_s = I ω₀
L_s = 5.20 x 10⁻³ x 2.87
L_s=0.015 kg.m²/s
initial angular momentum of the cockroach
L_c = - mvr
L_c = - 0.117 x 1.91 x 0.173
L_c = - 0.0387 kg.m²/s
total angular momentum of Both
L = 0.015 - 0.0387
L = - 0.0237 kg.m²/s
after cockroach stop inertia becomes
I_f = I + mr^2
I_f = 5.20 x 10⁻³+ 0.117 x 0.173^2
I_f = 8.7 x 10⁻³ kg.m²/s
final angular momentum of the disk
L_f = I_f ω_f
L_f = 8.7 x 10⁻³ x ω_f
using conservation of momentum
L_i = L_f
-0.0237 =8.7 x 10⁻³ x ω_f
[tex]\omega_f = \dfrac{-0.0237}{8.7 \times 10^{-3}}[/tex]
[tex]\omega_f = -2.72\ rad/s[/tex]
angular speed of Susan is [tex]\omega_f = -2.72\ rad/s[/tex]
The value is negative because it is in the opposite direction of cockroach.
[tex]|\omega_f |= 2.72\ rad/s[/tex]
b) the mechanical energy is not conserved because cockroach stopped in between.
The efficiency of a Stirling cycle depends on the temperatures of the hot and cold isothermal parts of the cycle.If you increase the upper temperature, keeping the lower temperature the same, does the efficiency increase,decrease, or remain the same?A. increaseB. decreaseC. remain the same
Answer:
A. increase
Explanation:
Stirling cycle having four processes
1.Two processes are constant temperature processes
2.Two processes are constant volume processes
The efficiency of Stirling cycle is same as the efficiency of Carnot cycle.
The efficiency of Stirling cycle given as
[tex]\eta=1-\dfrac{T_L}{T_H}[/tex]
[tex]T_L[/tex]=Lower temperature
[tex]T_H[/tex]=Upper temperature
If we increase then upper temperature while the lower temperature is constant then the efficiency of Stirling cycle will increase because the ratio of lower and upper temperature will decreases.
Therefore answer is
A. increase
The nearest known exoplanets (planets beyond the solar system) are around 20 light-years away. What would have to be the minimum diameter of an optical telescope to resolve a Jupiter-sized planet at that distance using light of wavelength 600 nm? (Express your answer to two significant figures.)
To solve the problem, it is necessary to apply the concepts related to the diffraction given in circular spaces. By definition it is expressed as
[tex]\Delta \theta = 1.22\frac{\lambda}{d}[/tex]
Where,
\lambda = Wavelength
d = Optical Diameter
[tex]\theta =[/tex] Angular resolution
In turn you can calculate the angle through the diameter and the arc length, that is,
[tex]\Delta \theta = \frac{x}{D}[/tex]
Where,
x = The length of the arc
D = Distance
From known data we know that Jupiter's diameter is,
[tex]x_J = 1.43*10^8m[/tex]
[tex]D = 20*9.4608*10^{15}[/tex]
[tex]\lambda = 600*10^{-9}m[/tex]
Replacing we have that,
[tex]\frac{x}{D} = 1.22\frac{\lambda}{d}[/tex]
[tex]\frac{1.43*10^8}{20*9.4608*10^{15} } = 1.22\frac{600*10^{-9}}{d}[/tex]
Re-arrange to find d,
[tex]d = 968.5m = 0.968Km[/tex]
Therefore the minimum diameter of an optical telescope to resolve a Jupiter-sized planet is 0.968Km.
A man fell out of an airplane and barely survived. He was moving at a speed of 100m/s just before landing in deep snow on a mountain side. Experts estimated that the average net force on him was 600 N as he plowed through the snow for 10 s. What was his mass, in kg
Answer:
Mass in kg will be 60 kg
Explanation:
We have given initial speed of the man u = 100 m/sec
And final speed v = 0 m/sec
Time is given, t = 10 sec
So acceleration [tex]a=\frac{v-u}{t}=\frac{0-100}{10}=-10m/sec^2[/tex]
Force is given , F= 600 N
From Newton's law we know that F = ma
So [tex]600=m\times 10[/tex]
mass = 60 kg
So mass in kg will be 60 kg
A person who weighs 685 N steps onto a spring scale in the bathroom, and the spring compresses by 0.88 cm. (a) What is the spring constant? N/m (b) What is the weight of another person who compresses the spring by 0.38 cm?
To solve this problem it is necessary to use the concepts of Force of a spring through Hooke's law, therefore,
[tex]F = kx[/tex]
Where,
k = Spring constant
x = Displacement
Initially our values are given,
[tex]F = 685N[/tex]
[tex]x = 0.88 cm[/tex]
PART A ) With this values we can calculate the spring constant rearranging the previous equation,
[tex]k = \frac{F}{x}[/tex]
[tex]k = \frac{685}{0.88*10^-2}[/tex]
[tex]k = 77840.9N/m[/tex]
PART B) Since the constant is unique to the spring, we can now calculate the force through the new elongation (0.38cm), that is
[tex]F = kx[/tex]
[tex]F = (77840.9)(0.0038)[/tex]
[tex]F = 295.79N[/tex]
Therefore the weight of another person is 265.79N.
Final answer:
To determine the spring constant (k) and the weight of another person, Hooke's law is applied. The spring constant is calculated using the weight of the first person and the distance the spring was compressed. Then, with the found spring constant, the weight for another compression distance is determined.
Explanation:
To solve both parts of the question, we make use of Hooke's law, which states that the force (F) needed to compress or extend a spring by some distance (x) is directly proportional to that distance. The law is usually formulated as F = kx, where k is the spring constant. To find the spring constant, we rearrange the formula to k = F/x.
Part A:We are given that a person weighing 685 N compresses the spring by 0.88 cm (which is 0.0088 m). Using the formula k = F/x, we plug in the values to get k = 685 N / 0.0088 m, which gives us the spring constant.
Part B: With the spring constant from part A, we can then determine the weight of another person by using the same Hooke's law. If the spring is compressed by 0.38 cm (which is 0.0038 m), we use the formula F = kx to calculate the new weight (force).
You’re in a mall and you need some money to buy a nose ring, but you’re broke. You decide to scoop some quarters out of the fountain. The water in the fountain is one foot deep. How far below the water do the quarters appear to be? The index of refraction of water is 4/3.
Answer: 0.75 ft
Explanation:
This situation is due to Refraction, a phenomenon in which a wave (the light in this case) bends or changes its direction when passing through a medium with an index of refraction different from the other medium. In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.
In addition, we have the following equation that states a relationship between the apparent depth [tex]{d}^{*}[/tex] and the actual depth [tex]d[/tex]:
[tex]{d}^{*}=d\frac{{n}_{1}}{{n}_{2}}[/tex] (1)
Where:
[tex]n_{1}=1[/tex] is the air's index of refraction
[tex]n_{2}=\frac{4}{3}=1.33[/tex] water's index of refraction.
[tex]d=1 ft[/tex] is the actual depth of the quarters
Now. when [tex]n_{1}[/tex] is smaller than [tex]n_{2}[/tex] the apparent depth is smaller than the actual depth. And, when [tex]n_{1}[/tex] is greater than [tex]n_{2}[/tex] the apparent depth is greater than the actual depth.
Let's prove it:
[tex]{d}^{*}=1 ft\frac{1}{1.33}[/tex] (2)
Finally we find the aparent depth of the quarters, which is smaller than the actual depth:
[tex]{d}^{*}=0.75 ft[/tex]
A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicular to the disk faces. A uniform force of 4.0 N is applied tangentially to the rim of the disk. What is the angular acceleration of the disk?
Final answer:
The angular acceleration of the disk is calculated using the torque exerted by the applied force and the moment of inertia of a uniform disk. The calculated angular acceleration is 5.33 rad/s².
Explanation:
The question involves calculating the angular acceleration of a rotating disk when a force is applied tangentially at its rim. To find the angular acceleration, we first need to calculate the torque exerted by the force and then use the moment of inertia of the disk to find the angular acceleration.
Given:
Mass of the disk (m) = 5.0 kgDiameter of the disk (d) = 30 cm = 0.30 m (radius r = d/2 = 0.15 m)Applied force (F) = 4.0 NThe torque (τ) exerted by the force is the product of the force and the radius at which it is applied (torque = force × radius), so:
τ = F × r = 4.0 N × 0.15 m = 0.60 N·m
The moment of inertia (I) for a uniform disk is given by (1/2)mr². Thus:
I = (1/2) × 5.0 kg × (0.15 m)² = 0.1125 kg·m²
Using the formula for angular acceleration (α = torque / moment of inertia), we get:
α = τ / I = 0.60 N·m / 0.1125 kg·m² = 5.33 rad/s²
Therefore, the angular acceleration of the disk is 5.33 rad/s².
A cannon fires a cannonball of mass 16.0 kg by applying a force of 2750 N along the 1.25 m length of the barrel. (a) How much work does the cannon do on the cannonball? (b) The cannon is aimed at a 25.0° angle above the horizontal. Assume gravity is the only other force acting on the cannonball as it moves through the cannon barrel. (That is, ignore all frictional forces.) What is the net work done by these 2 forces on the cannonball while it is in the cannon barrel?
To develop this problem it is necessary to apply the concepts related to Work and energy conservation.
By definition we know that the work done by a particle is subject to the force and distance traveled. That is to say,
[tex]W = F*d[/tex]
Where,
F= Force
d = Distance
On the other hand we know that the potential energy of a body is given based on height and weight, that is
[tex]PE = -mgh[/tex]
The total work done would be given by the conservation and sum of these energies, that is to say
[tex]W_{net} = W+PE[/tex]
PART A) Applying the work formula,
[tex]W = F*d\\W = 2750*1.25\\W = 3437.5J[/tex]
PART B) Applying the height equation and considering that there is an angle in the distance of 25 degrees and the component we are interested in is the vertical, then
[tex]PE = -mgh*sin25[/tex]
[tex]PE = -16*9.81*1.15sin25[/tex]
[tex]PE = -82.9J[/tex]
The net work would then be given by
[tex]W_{net} = W+PE[/tex]
[tex]W_{net} = 3437.5J-82.9J[/tex]
[tex]W_{net} = 3354.6J[/tex]
Therefore the net work done by these 2 forces on the cannonball while it is in the cannon barrel is 3355J
You are riding in an enclosed train car moving at 90 km/h. If you throw a baseball straight up, where will the baseball land?
a. In front of you.
b. In your hand.
c. Behind you.
d. Can't decide from the given information.
Answer:b. In your hand
Explanation: The objects from an outer perspective are moving at 90km/h all of them, even if there was a fly flying around in the train. Then if there is no change of speed of the train the moment the ball leaves the hand its relative speed with respect to the hand is zero (0), then it will land in the hand again. So the only way for it not to happen is either you move the hand or the train changes its speed, therefore the relative speed of the
ball with respect to the hand would be differente from zero and further information would be required, but since it is not stated that neither the train accelerates nor the hand is going to be moved, we can affirm the ball will land in the hand.
A fisherman’s scale stretches 3.3 cm when a 2.6-kg fish hangs from it. The fish is pulled down 2.5 cm more and released so that it oscillates up and down with a simple harmonic motion (SHM). ? (a) What is the spring stiffness constant? (b) What is be the amplitude? (c) What is the period and frequency of oscillations?
a) The spring constant is 772.1 N/m
b) The amplitude is 0.025 m
c) The period is 0.365 s, the frequency is 2.74 Hz
Explanation:
a)
At equilibrium, the weight of the fish hanging on the spring is equal to the restoring force of the spring. Therefore, we can write:
[tex]mg=kx[/tex]
where
(mg) is the weight of the fish
kx is the restoring force
m = 2.6 kg is the mass of the fish
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
k is the spring constant
x = 3.3 cm = 0.033 m is the stretching of the spring
Solving for k, we find:
[tex]k=\frac{mg}{x}=\frac{(2.6)(9.8)}{0.033}=772.1 N/m[/tex]
b)
Later, the fish is pulled down by 2.5 cm (0.025 m), and the system starts to oscillate.
The amplitude of a simple harmonic motion is equal to the maximum displacement of the system. In this case, when the fish is pulled down by 2.5 cm and then released, immediately after the releasing the fish moves upward, until it reaches the same displacement on the upper side (+2.5 cm).
This means that the amplitude of the motion corresponds also to the initial displacement of the fish when it is pulled down: therefore, the amplitude is
A = 2.5 cm = 0.025 m
c)
The period of oscillation of a mass-spring system is given by
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
where
m is the mass
k is the spring constant
Here we have
m = 2.6 kg
k = 772.1 N/m
So, the period is
[tex]T=2\pi \sqrt{\frac{2.6}{772.1}}=0.365 s[/tex]
And the frequency is given by the reciprocal of the period, therefore:
[tex]f=\frac{1}{T}=\frac{1}{0.365}=2.74 Hz[/tex]
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A bicycle is traveling north at 5.0 m/s. The mass of the wheel, 2.0 kg, is uniformly distributed along the rim, which has a radius of 20 cm. What are the magnitude and direction of the angular momentum of the wheel about its axle?
Answer:2
Explanation:
Given
Velocity of bicycle is 5 m/s towards north
radius of rim [tex]r=20 cm[/tex]
mass of rim [tex]m=2 kg[/tex]
Angular momentum [tex]\vec{L}=I\cdot \vec{\omega }[/tex]
[tex]I=mr^2=2\times 0.2^2=0.08 kg-m^2[/tex]
[tex]\omega =\frac{v}{r}=\frac{5}{0.2}=25 rad/s[/tex]
[tex]L=0.08\times 25=2kg-m^2/s[/tex]
direction of Angular momentum will be towards west
A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the Earth. If we ignore atmospheric friction, with what speed does the object strike the surface of the earth? (G = 6.67 × 10-11 N · m2/kg2, MEarth = 5.97 × 1024 kg, the polar radius of the earth is 6357 km)
In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,
[tex]KE = \fract{1}{2}mv^2[/tex]
[tex]PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))[/tex]
Where,
M = Mass of Earth
m = Mass of Object
v = Velocity
r = Radius
G = Gravitational universal constant
Our values are given as,
[tex]m = 910 Kg[/tex]
[tex]r_1 = 1200 + 6371 km = 7571km[/tex]
[tex]r_2 = 6371 km,[/tex]
Replacing we have,
[tex]\frac{1}{2} mv^2 = -GMm(\frac{1}{r_1}-\frac{1}{r_2})[/tex]
[tex]v^2 = -2GM(\frac{1}{r_1}-\frac{1}{r_2})[/tex]
[tex]v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})[/tex]
[tex]v = 4456 m/s[/tex]
Therefore the speed of the object when striking the surface of earth is 4456 m/s
The speed of a 910-kg object striking the Earth after falling from 1200 km above the North Pole can be found using conservation of mechanical energy and the formula for gravitational potential energy.
Explanation:To determine the speed at which a 910-kg object strikes the surface of the Earth when released from rest at an altitude of 1200 km above the North Pole, we can use the conservation of mechanical energy. The total mechanical energy (kinetic plus potential) at the initial and final points must be equal since no external work is done on the system (we ignore atmospheric friction).
Initially, the object has potential energy due to its altitude above Earth and no kinetic energy since it's at rest. When it strikes the Earth, it has kinetic energy and minimal potential energy. Setting the initial and final total energies equal gives:
Ui + Ki = Uf + Kf,
where U is potential energy and K is kinetic energy. The potential energy can be calculated using the formula U = -GMEarthm/r, where m is the object's mass, r is the distance from the object to the center of Earth, and G is the gravitational constant. The kinetic energy is given by K = (1/2)mv2, where v is the velocity of the object.
The initial distance ri to the center of Earth is 6357 km + 1200 km (release altitude), and when the object strikes Earth, rf = 6357 km (polar radius of Earth). We know Ki = 0 and Uf is small enough to be negligible at the Earth's surface.
By plugging in the values, solving the equation for v, and taking the square root, we find the speed of the object when it strikes the ground.