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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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3 years ago

3. A fighter jet covers a distance of 895m while accelerating from 22m/s to 35m/s. How long (in

larisa86 [58]

Answer:

31.404 seconds

Explanation:

To answer this equation, SUVAT is your best option utilizing and rearranging the known values to solve for the unknown.

here we have the values for

s=895

u=22

v=35

t= the unknown value

in this instant the equation s=0.5 x (u+v)t is the best equation to use

so we sub in the known values

895=0.5 x (22+35)t

rearrange to solve for t

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3 years ago

A diode vacuum tube consists of a cathode and an anode spaced 5-mm apart. If 300 V are applied across the plates. What is the ve

MAXImum [283]

Answer:

Explanation:

There is electric field between the plates whose value is given by the following expression

electric field E = V /d where V is potential between the plates and d is distance between them

E = 300 / 5 x 10⁻³

= 60 x 10³ N/c

Force on electron = q E where q is charge on the electron

F = 1.6 X 10⁻¹⁹ X 60 X 10³ = 96 X 10⁻¹⁶ N.

Acceleration a = force / mass

a = 96 x 10⁻¹⁶/ mass = 96 x 10⁻¹⁶ / 9.1 x 10⁻³¹

= 10.55 x 10¹⁵ m / s²

For midway , distance travelled

s = 2.5 x 10⁻³ m

s = 1\2 a t²

t = $\sqrt{\frac{2s}{a\\ } }$

= $\sqrt{\frac{2\times2.5\times10^{-3}}{ 10.55\times10^{15}}$

t = .474 x 10⁻¹⁸ s

For striking the plate time is calculated as follows

t = $[tex]\sqrt{\frac{2\times5\times10^{-3}}{ 10.55\times10^{15}}$ [/tex]

t = 0.67 x 10⁻¹⁸ s

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3 years ago

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Answer:

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3 years ago