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faust18 [17]
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?
Physics
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

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12000 inches to yards
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ANSWER

\begin{equation*} 333.33\text{ yds} \end{equation*}

EXPLANATION

We want to convert 12000 inches to yards.

To do this, divide the value in inches by 36:

\begin{gathered} 1\text{ in }=\frac{1}{36}\text{ yd} \\  \\ 12000\text{ in }=\frac{12000}{36}\text{ yds }=333.33\text{ yds} \end{gathered}

That is the answer.

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4 years ago
A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

6 0
3 years ago
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