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3 years ago

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Galaxy B moves away from galaxy A at 0.577 times the speed of light. Galaxy C moves away from galaxy B in the same direction at

0.731 times the speed of light. How fast does galaxy C recede from galaxy A?

1 answer:

Pepsi [2]3 years ago

7 0

Answer:

The value is $p = 0.7556 c$

Explanation:

From the question we are told that

The speed at which galaxy B moves away from galaxy A is $v = 0.577c$

Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

The speed at which galaxy C moves away from galaxy B is $u = 0.731 c$

Generally from the equation of relative speed we have that

$u = \frac{p - v}{ 1 - \frac{ p * v}{c^2} }$

Here p is the velocity at which galaxy C recede from galaxy A so

$0.731c = \frac{p - 0.577c }{ 1 - \frac{ p * 0.577c}{c^2} }$

=> $0.731c [1 - \frac{ p * 0.577}{c}] = p - 0.577c$

=> $0.731c - 0.4218 p = p - 0.577c$

=> $0.731c + 0.577c = p + 0.4218 p$

=> $1.308 c = 1.731 p$

=> $p = 0.7556 c$

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3 years ago

Can you think of a scenario when the kinetic and gravitational potential energy could both be zero ? Describe or draw how this c

Inga [223]

Both kinetic and gravitational potential energy can become zero at infinite distance from the Earth.

Consider an object of mass <em>m </em>projected from the surface of the Earth with a velocity <em>v. </em>

The total energy of the body on the surface of the Earth is the sum of its kinetic energy $\frac{1}{2} mv^2$ and gravitational potential energy $-\frac{GMm}{R^2}$ .

here, <em>M</em> is the mass of the Earth, <em>R</em> is the radius of Earth and <em>G</em> is the universal gravitational constant.

The gravitational potential energy of the object is negative since it is in an attractive field, which is the gravitational field of the Earth.

The energy of the object on the surface of the earth is given by,

$E_i=\frac{1}{2} mv^2-\frac{GMm}{R^2}$

As the object rises upwards, it experiences deceleration due to the gravitational force of the Earth. Its velocity decreases and hence its kinetic energy decreases.

The decrease in kinetic energy is manifested as an equal increase in potential energy. The potential energy becomes less and less negative as more and more kinetic energy is converted into potential energy.

At a height <em>h</em> from the surface of the Earth, the energy of the object is given by,

$E_h=\frac{1}{2} mv_h^2-\frac{GMm}{(R+h)^2}$

The velocity $v_h$ is less than <em>v</em>.

When h =∞, the gravitational potential energy increases from a negative value to zero.

If the velocity of projection is adjusted in such a manner that the velocity decreases to zero at infinite distance from the earth, the object's kinetic energy also becomes equal to zero.

Thus, it is possible for both kinetic and potential energies to be zero at infinite distance from the Earth. In this case, kinetic energy decreases from a positive value to zero and the gravitational potential energy increases from a negative value to zero.

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3 years ago

A motorcyclist drives around a bend with a 20 m radius, with a constant velocity of 3 m/s. The motorcyclist and the motorcycle h

Pavel [41]

Answer:

$a=0.45\ m/s^2$

Explanation:

Given that,

The radius of a bend, r = 20 m

Velocity of motorcyclist, v = 3 m/s

The combined mass of motorcyclist and the motorcycle is 50 kg

We need to find the motorcyclist’s centripetal acceleration. The formula used to find the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(3)^2}{20}\\\\a=0.45\ m/s^2$

So, the acceleration of the motorcyclist is $0.45\ m/s^2$ .

3 0

3 years ago

An automobile engine delivers 55.0 hp. How much time will it take for the engine to do 6.22 × 105 J of work? One horsepower is e

Gennadij [26K]

Answer:

15.2 s

Explanation:

Convert hp to W:

55.0 hp × 746 W/hp = 41,030 W

Power = energy / time

41030 W = 6.22×10⁵ J / t

t = 15.2 s

8 0

2 years ago

A piece of steel is 11.5cm long at 22C. It is heated to 1221C, close to its melting point. How long is it, in cm, at the high te

Nataly [62]

Answer:

The length at the final temperature is 11.7 cm.

Explanation:

We need to use the thermal expansion equation:

$\Delta L=\alpha L_{0}\Delta T$

Where:

L(0) is the initial length
ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
ΔL is the final length minus the initial length (L(f)-L(0))
α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)

So, we have:

$L_{f}-L_{0}=\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=L_{0}+\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=0.115+(12.5*10^{-6})(0.115)(1221-22)$

$L_{f}=0.117\: m$

Therefore, the length at the final temperature is 11.7 cm.

I hope it helps you!

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2 years ago