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I am Lyosha [343]
2 years ago
11

What does Newton's 2nd Law explain?

Physics
1 answer:
kow [346]2 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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Answer:

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Explanation:

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3 years ago
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Which statement describes potential energy?
gayaneshka [121]
I believe the answer is stored energy.
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3 years ago
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If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th
mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

x=ut

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

y=\frac{1}{2} gt^2

Substitute 9.8 m/s² for g and 0.461 s for t.

y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by,(2.0 m)-(1.04 m)=0.96 m

The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>

8 0
3 years ago
A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20 m/s by a 5620 N braking force actin
Dennis_Churaev [7]

Answer:

the distance traveled by the car is 42.98 m.

Explanation:

Given;

mass of the car, m = 2500 kg

initial velocity of the car, u = 20 m/s

the braking force applied to the car, f = 5620 N

time of motion of the car, t = 2.5 s

The decelaration of the car is calculated as follows;

-F = ma

a = -F/m

a = -5620 / 2500

a = -2.248 m/s²

The distance traveled by the car is calculated as follows;

s = ut + ¹/₂at²

s = (20 x 2.5) + 0.5(-2.248)(2.5²)

s = 50 - 7.025

s = 42.98 m

Therefore, the distance traveled by the car is 42.98 m.

5 0
2 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
puteri [66]

Answer:

t=5.3687\ s  is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is 0\ mph\ to\ 70\ mph.

  • Initial velocity of the car, v_i=0\ mph
  • final velocity of the car during the test, v_f=32\ mph=14.3052\ m.s^{-1}
  • Time taken to accelerate form zero to 32 mph at full power, t=1.2\ s
  • initial velocity of the car, u_i=0\ mph
  • final desired velocity of the car, u_f=64\ mph=28.6105\ m.s^{-1}

Now the acceleration of the car:

a=\frac{v_f-v_i}{t}

a=\frac{14.3052-0}{1.2}

a=11.921\ m.s^{-1}

Now using the equation of motion:

u_f=u_i+a.t

64=0+11.921\times t

t=5.3687\ s is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0
3 years ago
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