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andre [41]
3 years ago
8

A 1,200 kg car travels at 20 m/s. what is it’s momentum ?

Physics
1 answer:
WINSTONCH [101]3 years ago
6 0

Answer:c  vx v xv xv x

Explanation:

mlmcmcm

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Use this formula to solve this problem:
Maslowich
Well, you gave us the formula to calculate power from work and time,
but you didn't give us the formula for work.  We have to know that.

             Work = (force) x (distance)

The work to raise Sara to the top of the hill is

               Work = (300 N) x (15 meters)

                         =   4,500 newton-meters  =  4,500 joules .

Now we're ready to use the formula that you gave us.  (Thank you.)

                Power = (work) / (time)

                            = (4,500 joules) / (10 seconds)

                            450 joules/second  =  450 watts.       
6 0
3 years ago
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A weight lifter lifts a dumbbell a certain height in 2.0 s, while a competitor does the same workin 1.0 s. Compared to the power
Aloiza [94]

Answer:

a. one-half as great

Explanation:

The power developed by the first lifter is one-half as great as that of the second person.

  Power is defined as the rate at which work is done;

          Power  = \frac{workdone}{time}

Since the two lifters do the same work at different time, let us estimate their power;

       P₁ = \frac{workdone}{2}                     P₂ = \frac{workdone }{1}

   We see that for P₁, power is half of the work done whereas in P₂ power is the same as the work done.

Therefore,

           The power of the first weight lifter is one-half the second lifter.

4 0
2 years ago
An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X
Verizon [17]

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

Speed of rocket 1 with respect to rocket 2 :

$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$

$u' = \frac{0.7 c}{1.12}$

u'=0.625 c

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

5 0
3 years ago
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