State whether the following statement are true or false .

An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi

Arturiano [62]

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

$V_x'=0.8c\\V=0.6c\\m=4*10^5kg$

Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

$KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}$

So, to $V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016$ find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
We have an expression from Lorents transformation for relativistic law of addition of velocities as,

$V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V$

Substituting values, we get,

$V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c$

Thus, the KE will be,

$KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J$

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

3 0

1 year ago

A ball accelerates uniformly at +4.0 m/s2

kirill115 [55]

Answer:

483883838939384847848484

8 0

2 years ago

true or false? short-wave heat radiation is given off by the earth when the earth absorbs long-wave radiation.

Greeley [361]

Answer:

Hope this helps u, pls mark me brainlist

Explanation:

The sun emits shortwave radiation because it is extremely hot and has a lot of energy to give off. Once in the Earth's atmosphere, clouds and the surface absorb the solar energy. The ground heats up and re-emits energy as longwave radiation in the form of infrared rays.

4 0

2 years ago

At a stoplight, a truck traveling at 15 m/s passes a car as it starts from rest. The truck travels at constant velocity and the

Art [367]

The time that the car take to catch up to truck would be :

1/2 ar^2 = vt

1/2 (3) t^2 = 15t

1.5 t^2 = 15t

t ^2 = 10s

Hope this helps

4 0

3 years ago

Modeled after the two satellites problem (slide 12 in Lecture13, Universal gravity), imagine now if the period of the satellite

ikadub [295]

Answer:

R = 6.3456 10⁴ mile

Explanation:

For this exercise we will use Newton's second law where force is gravitational force

F = m a

The satellite is in a circular orbit therefore the acceleration is centripetal

a = v² / r

Where the distance is taken from the center of the Earth

G m M / r² = m v² / r

G M / r = v²

The speed module is constant, let's use the uniform motion relationships, with the length of the circle is

d = 2π r

v = d / t

The time for a full turn is called period (T)

Let's replace

G M / r = (2π r / T)²

r³ = G M T²²2 / 4π²

r = ∛ (G M T² / 4π²)

We have the magnitudes in several types of units

T = 88.59 h (3600 s / 1h) = 3.189 10⁵ s

Re = 6.37 10⁶ m

Let's calculate

r = ∛ (6.67 10⁻¹¹ 5.98 10²⁴ (3,189 10⁵)²/4π²)

r = ∛ (1.027487 10²⁴)

r = 1.0847 10⁸ m

This is the distance from the center of the Earth, the distance you want the surface is

R = r - Re

R = 108.47 10⁶ - 6.37 10⁶

R = 102.1 10⁶ m

Let's reduce to miles

R = 102.1 10⁶ m (1 mile / 1609 m)

R = 6.3456 10⁴ mile

6 0

2 years ago