Ask question

Ask question

All categories

3 years ago

11

Two rockets, A and B, approach the earth from opposite directions at speed 0.80c. The length of each rocket measured in its rest

frame is 100 m. What is the length of rocket A as measured by the crew of rocket B?

1 answer:

Fantom [35]3 years ago

4 0

Answer:

The contracted length as measured is 22.22 m

Solution:

As per the question:

Speed of the rockets, v = 0.8c

Length of the rocket, L = 100 m

Now,

To calculate the length of rocket A w.r.t rocket B:

The relative speed of the rockets A and B approaching the earth from the opposite directions:

$Relative speed,\ v_{r} = \frac{- 0.8c - 0.8c}{1 + \frac{v^{2}}{c^{2}}}$

$v_{r} = \frac{- 1.6c}{1 + \frac{(0.8c)^{2}}{c^{2}}} = 0.975c$

The contracted length is given by:

$L' = \frac{L}{\sqrt{1 - \frac{v_{r}^{2}}{c^{2}}}}$

$L' = \frac{100}{\sqrt{1 - \frac{(0.975c)^{2}}{c^{2}}}} = 22.22\ m$

You might be interested in

Proved that<br>V = u+at<br>

kondaur [170]

Answer:

$\sf Proof \ below$

Explanation:

We know that acceleration is change in velocity over time.

$\sf a=\frac{\triangle v}{t}$

$\sf a=\frac{v-u}{t}$

v is the final velocity and u is the initial velocity.

Solve for v.

Multiply both sides by t.

$\sf at=v-u$

Add u to both sides.

$\sf at + u=v$

3 0

3 years ago

Read 2 more answers

Which statement best explains why radon and krypton do not bond easily with other elements?

mote1985 [20]

Answer.B

Explanation:

6 0

3 years ago

Read 2 more answers

An object with height h, mass M, and a uniform cross-sectional area A floats upright in a liquid with density ρ.

soldi70 [24.7K]

** Missing information: The vertical distance from surface of liquid to bottom of the object is sought in this question, with the condition that the object is at equilibrium **

Ans: The vertical distance = y = M/(ρA)

Explanation:

Support the vertical distance = y

Object's density = M/(A*h) (since A*h = volume)

By applying the condition,

(M/(Ah))/ρ = y/h

M/(ρAh) = y/h

y = M/(ρA)

7 0

3 years ago

A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an

Nat2105 [25]

Answer:

a) $T=0.0031416s$

b) $v_{max}=6.283\frac{m}{s}$

c) $E=0.1974J$

d) $F=80N$

e) $F=40N$

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by $X=Acos(\omega t +\phi)$ (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

$\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)$ (2)

For the acceleration

$\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)$ (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

$A\omega^2=8x10^{3}\frac{m}{s^2}$

Since we know the amplitude A=0.002m we can solve for $\omega$ like this:

$\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}$

And we with this value we can find the period with the following formula

$T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s$

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

$v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}$

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

$E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J$

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens $cos(\omega t +\phi)=1$ , and we also know that the maximum acceleration is given by::

$|\frac{d^2X}{dt^2}|=A\omega^2$

So then we have that:

$F=ma=mA\omega^2$

And since we have everything we can find the force

$F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N$

6) Part e

When the mass it's at the half of it's maximum displacement the term $cos(\omega t +\phi)=1/2$ and on this case the acceleration would be given by;

$|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}$

And the force would be given by:

$F=ma=\frac{1}{2}mA\omega^2$

And replacing we have:

$F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N$

8 0

2 years ago

Keith_Richards [23]

The answer is C) the density of the rock

Density of rock is the dependent variable, because it depends on the temperature. The temperature can’t be the dependent variable because ,the density of a rock does not have magical powers that can change temperature of a room. However changing the temperature of the room ,will change the density of the rock. Hope this helps !

4 0

3 years ago