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I am Lyosha [343]
3 years ago
11

What does Newton's 2nd Law explain?

Physics
1 answer:
kow [346]3 years ago
6 0

Answer:

C. Why you must push harder to move a car farther.

Explanation:

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

Acceleration = \frac {Force}{mass}

Hence, Newton's 2nd Law explains why you must push harder to move a car farther because of its mass. Thus, it is important to increase the force that the engine provides and decrease the mass of the car.

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If a girl is standing still and holding a box, is she doing any work? (No)
SpyIntel [72]

Answer:

gravity

Explanation:

7 0
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Average velocity is different than average speed because calculating average velocity involves a)Distance b)Time c)Motion d)Disp
likoan [24]

Answer:

The answer is D.

Explanation:

Average speed involve just distance and time but average velocity includes displacement and time.

(Correct me if I am wrong)

6 0
3 years ago
Will an object with the density of 0.66g float?
Firdavs [7]
Yes yes it will............
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In a pool game, the cue ball, which has an initial speed of 3.0 m/s, make an elastic collision with the eight ball, which is ini
zhenek [66]

Explanation:

Given

initial speed(u)=3 m/s

mass of each ball is m

Let the cue ball is moving in x direction initially

In elastic collision Energy and momentum is conserved

Let u be the initial velocity and v_1 , v_2 be the final velocity of 8 ball and cue ball respectively

\frac{mu^2}{2}+0=\frac{mv^2_1}{2}+\frac{mv^2_2}{2}

The angle after which cue ball is deflected is given by

\theta _1=90-40=50^{\circ}

Conserving momentum in x direction

mu=mv_1cos40+mv_2cos50

3=v_1cos40+v_2cos50

Along Y axis

0+0=v_1sin40-v_2sin50

v_1sin40=v_2sin50

substitute the value of v_1

we get v_2=1.912 m/s

v_1=2.27 m/s

5 0
3 years ago
A cannonball is fired on flat ground
algol [13]

hmax = 5740.48 m. The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.

This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.

V₀ = 420m/s and θ₀ = 53.0°

So, when the cannonball is fired it has horizontal and vertical components:

V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s

V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s

When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:

Vy = V₀y - g tₐ = 0

tₐ = V₀y/g

tₐ = (335.43m/s)/(9.8m/s²) = 34.23s

Then, the maximum height is reached in the instant tₐ = 34.23s:

h = V₀y tₐ - 1/2g tₐ²

hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²

hmax = 11481.77m - 5741.29m

hmax = 5740.48m

3 0
3 years ago
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