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3 years ago

Which of the following is true of changes in momentum?

Physics

1 answer:

soldi70 [24.7K]3 years ago

6 0

Answer: ⇒ Answer is 3

<h2>Explanation: momentum = mass × velocity</h2>

"A small force may produce a large change in momentum by acting on a very massive object".

THEY HAVEN'T GIVEN US THE TIME PERIOD NOR THE DISTANCE TRAVELED. THEREFORE WE CANNOT ACTUALLY DECIDE IF THE FORCE IS KEPT FOR A LONG TIME OR SHORT TIME. ANYWAYS SINCE THE MASS IS GIVEN AS MASSIVE , THE MOMENTUM SHOULD BE DEFINITELY HIGH.

WHY I SAY OTHERS ARE WRONG:

1) For a small force to give a large change in momentum, it should act for a long time interval.

2) By applying a large force for a short time interval, the change of momentum should be large.

3) Correct answer.

4) Acting over a short distance can be the same as acting over a short period of time.Therefore the distance should be large in order for a larger momentum.

I HOPE IT HELPS!

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svlad2 [7]

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3 years ago

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3 years ago

A 3.0 kg block is pushed by a 14 N force. If µ = 0.6, will the block move?

Anna71 [15]

Answer:

The block will not move.

Explanation:

We'll begin by calculating the frictional force. This can be obtained as follow:

Coefficient of friction (µ) = 0.6

Mass of block (m) = 3 Kg

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (R) = mg = 3 × 10 = 30 N

Frictional force (Fբ) =?

Fբ = µR

Fբ = 0.6 × 30

Fբ = 18 N

From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.

4 0

3 years ago

If the force squeezing two surfaces together is decreased, the force of dry sliding friction between the two surfaces will most

Anastasy [175]

If the force squeezing two surfaces together is decreased, the force of dry sliding friction between the two surfaces will most likely decrease.

therefore your answer is B)decrease

Hope this helps you- have a good day bro cya)

7 0

2 years ago

The earth has a vertical electric field at the surface, pointing down, that averages 119 N/C . This field is maintained by vario

velikii [3]

Answer:

$q=5.37*10^{5}C$

Explanation:

If we assume that the Earth is a spherical conductor, according to Gauss's Law, the electric field is given by:

$E=\frac{kq}{r^2}$

Here k is the Coulomb constant, the excess charge on the Earth's surface and r its radius. Solving for q:

$q=\frac{Er^2}{k}\\q=\frac{119\frac{N}{C}(6.371*10^6m)^2}{8.99\frac{N\cdot m^2}{C^2}}\\q=5.37*10^{5}C$

5 0

3 years ago