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gregori [183]
3 years ago
12

What is momentum?????​

Physics
1 answer:
Leokris [45]3 years ago
7 0

Momentum is defined as the mass (m) times the velocity (v). If an object is steady so its velocity is zero resulting in zero momentum.

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5 points
Tresset [83]

Answer:

\frac{{v}^{2} g}{2 }

8 0
2 years ago
Please help me plsss!​
alexira [117]

Answer:

A

Explanation:

3 0
2 years ago
At what point on the position-time graph shown is the object's instantaneous velocity greatest?
ioda

I'm probably going to have to say C. E as it seems the steepest right around there. If I'm wrong on that, it has to be B. B

7 0
3 years ago
Read 2 more answers
What is the strength of an electric field that will put a force of<br> 1.28 x 10-15 N on a proton?
PilotLPTM [1.2K]

Answer: E =  7,490.6 N/C

Explanation:

If we have a field E, and a particle with a charge q, the force that the particle experiences is:

F = E*q

In this case, we know that the force is:

F = 1.2*10^(-15) N

And we know that the particle is a proton, where the charge of a proton is:

q = 1.602*10^(-19) C

Then we can replace these two values in the equation to get:

1.2*10^(-15) N = E*1.602*10^(-19) C

We just need to isolate E.

(1.2*10^(-15) N)/(1.602*10^(-19) C) = E

7,490.6 N/C = E

That is the strength of the electric field.

8 0
3 years ago
Can someone pleaseeee answer this !!!!!!
LenaWriter [7]

Answer:

The person with locked legs will experience greater impact force.

Explanation:

Let the two persons be of nearly equal mass (say m)

The final velocity of an object (person) dropped from a height H (here 2 meters) is given by,

v=\sqrt{2gH}

(g = acceleration due to gravity)

which can be derived from Newton's equation of motion,

v^2=u^2+2aS

Now, the time taken (say t ) for the momentum ( mv ) to change to zero will be more in the case of the person who bends his legs on impact than who keeps his legs locked.

We know that,

Force=\frac{\Delta(mv)}{t}

Naturally, the person who bends his legs will experience lesser force since t is larger.

3 0
3 years ago
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