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Alex_Xolod [135]
1 year ago
7

While in Earth's orbit, an 80-kg astronaut carrying a 20-kg tool kit is initially drifting toward a stationary space shuttle at

a speed of 2 m/s. If she throws the tool kit toward the shuttle with a speed of 6 m/s, her final speed is what value
Physics
1 answer:
aleksklad [387]1 year ago
8 0

The answer is 1 m/s toward the shuttle.

In physics, conservation means that the initial and final values are equivalent. This implies that a system's total starting and total final momentums will be identical in terms of momentum. The force acting on an item will always be equal to the change in the object's momentum over time, according to Newton's second law.

Therefore, initial momentum = final momentum

                               Pi           =         Pf

                           100*2         =        20*6+80*v

                            200           =         120+80*v

                          200-120      =          80*v

                              80           =          80*v

                               1            =           v

                               v            =           1 m/s

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Read 2 more answers
10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when
SVEN [57.7K]

Answer:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

Explanation:

For this case  we can use the second law of Newton given by:

\sum F = ma

The friction force on this case is defined as :

F_f = \mu_k N = \mu_k mg

Where N represent the normal force, \mu_k the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

F_f = ma

Replacing the friction force we got:

\mu_k mg = ma

We can cancel the mass and we have:

a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}

And now we can use the following kinematic formula in order to find the distance travelled:

v^2_f = v^2_i - 2ad

Assuming the final velocity is 0 we can find the distance like this:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

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