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3 years ago

A ball is released from rest at a height of 10 m and falls freely to the

Physics

1 answer:

ElenaW [278]3 years ago

6 0

Answer:

The new kinetic energy would be 16 times greater than before.

Explanation:

Kinetic energy is found using this formula:

KE = 1/2mv²
where KE = kinetic energy (J), m = mass (kg), and v = velocity (m/s)

We can see that kinetic energy is directly proportional to the square of the velocity, meaning that if the speed was increased by 4 times, then the kinetic energy would get increased by a factor of 16.

The velocity just before the ball hits the ground can be found by the equation:

√(2gh)

Let's substitute h = 10 m and h = 40 m into this formula.

√(2g(10))
√(2g(40))

We can see that the velocity increases by a factor of 4 (10 m → 40 m).

Therefore, this means that the kinetic energy would also be increased by a factor of (4)² = 16. Thus, the answer is D) The new kinetic energy would be 16 times greater than before.

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3 years ago

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3 years ago

A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s' for 10 secon

MArishka [77]

Answer:

Distance of 400m.

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Use your kinematics equation to solve for distance (we can use kinematics b/c acceleration is constant).

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2 years ago

Two balls are thrown against a wall. Ball 1 has a much higher speed than ball 2.

Sunny_sXe [5.5K]

Let both the balls have the same mass equals to m.

Let $v_1$ and $v_2$ be the speed of the ball1 and the ball2 respectively, such that

$v_1>v_2\;\cdots(i)$

Assuming that both the balls are at the same level with respect to the ground, so let h be the height from the ground.

The total energy of ball1= Kinetic energy of ball1 + Potential energy of ball1. The Kinetic energy of any object moving with speed, $v$ , is $\frac 12 m v^2$

and the potential energy is due to the change in height is $mgh$ [where $g$ is the acceleration due to gravity]

So, the total energy of ball1,

$=\frac 12 m v_1^2 + mgh\;\cdots(ii)$

and the total energy of ball1,

$=\frac 12 m v_2^2 + mgh\;\cdots(iii)$ .

Here, the potential energy for both the balls are the same, but the kinetic energy of the ball1 is higher the ball2 as the ball1 have the higher speed, refer equation (i)

So, $\frac 12 m v_1^2 >\frac 12 m v_2^2$

Now, from equations (ii) and (iii)

The total energy of ball1 hi higher than the total energy of ball2.

6 0

3 years ago