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natka813 [3]
2 years ago
11

What is an athlete’s momentum if the athlete has a velocity of 7.5 m/s and has a mass of 70kg?

Physics
1 answer:
Bezzdna [24]2 years ago
8 0

Answer:

525 kg.m/s

Explanation:

★ Momentum = Mass× Velocity

→ P = (7.5 × 70) kg.m/s

→ P = (75 × 7) kg.m/s

→ <u>P</u><u> </u><u>=</u><u> </u><u>5</u><u>2</u><u>5</u><u> </u><u>kg</u><u>.</u><u>m</u><u>/</u><u>s</u>

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Answer:

Explanation:

The acceleration of the ball would be due to the downward force of gravity, 9.8m/s^2. In order to find the displacement given that interval of time, you have to use the corresponding kinematic formula:

d=v_it+at^2/2

The initial velocity was given, the time was given, and the acceleration was given. Therefore:

d=(22m/s)(4s)+(9.8m/s^2)(4s)^2/2

d=166.4m

To find the required time given a desired final velocity, we can use:

v_f=v_i+at

60m/s=22m/s+(9.8m/s^2)(t)

38=9.8t

t=3.9s

6 0
2 years ago
Can someone help me with this please
andrezito [222]
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3 0
3 years ago
When we do dimensional analysis, we do something analogous to stoichiometry, but with multiplying instead of adding. Consider th
Ksivusya [100]

Answer:

The  dimension is  D =  L ^{2} T^{-1}

Explanation:

From the question we are told that

     J  =  -D \frac{dn}{dx}

Here  [J] = \frac{1}{L^2 T}

       [n] =\frac{1}{L^3}

        [x] = L

So

    \frac{1}{L^2 T} =  -D \frac{d(\frac{1}{L^3})}{d[L]}

Given that the dimension represent the unites of  n and  x then the differential  will not effect on them

So

\frac{1}{L^2 T} =  -D \frac{(\frac{1}{L^3})}{[L]}

=>   D =  \frac{L^{-2} T^{-1} * L }{L^{-3}}

=>   D =  L ^{2} T^{-1}

   

5 0
3 years ago
How many Medals in the Luge does the U.S. have?
Dafna11 [192]

Answer:

American athletes have won a total of 2,523 medals (1,022 of them gold) at the Summer Olympic Games and another 305 (105 of them gold) at the Winter Olympic Games, making the United States the most prolific medal-winning nation in the history of the Olympics.

Explanation:

Hope this helps! ^^

7 0
3 years ago
Which are ways to improve the design of this experiment? Check all that apply.
Arte-miy333 [17]

Answer:

* Experiment with a higher range of materials

* Use a galvanometer.

* Vary in number of coils of the electromagnet

Explanation:

This is an experiment of electricity and magnetism, in general the best way to improve the results are:

* Experiment with a higher range of materials

   allowing to know the scope of the initial assumptions

* Use a galvanometer.

  The more accurate the readings the error of the derived quantities is the less which will improve the precision of the experiment.

* Vary in number of coils of the electromagnet

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3 0
3 years ago
Read 2 more answers
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