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suter [353]
2 years ago
13

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

Physics
1 answer:
julsineya [31]2 years ago
3 0

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

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andrew11 [14]

Given :

A bicycle racer rides from a starting marker to a turnaround marker at 10 m/s.

She then rides back along the same route from the turnaround marker to the starting marker at 25 m/s.

To Find :

Her average speed for the whole race.

Solution :

We know, average velocity is given by :

Average \ Velocity = \dfrac{v_1 + v_2}{2}\\\\Average\ Velocity = \dfrac{10 + 25}{2}\\\\Average \ Velocity = \dfrac{35}{2}\\\\Average \ Velocity = 17.5 \ m/s

Therefore, average speed for the whole race is given by 17.5 m/s.

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fgiga [73]
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3 years ago
The inductor in the RLC tuning circuit of an AM radio has a value of 250 mHmH . You may want to review (Pages 857 - 860) . Part
ANEK [815]

Answer:

The value of variable capacitor is 1.89 \times 10^{-13} F

Explanation:

Given :

Inductance L = 250 \times 10^{-3} H

Frequency f = 731 \times 10^{3} Hz

According to the cutoff frequency,

   f = \frac{1}{2\pi \sqrt{LC} }

Now we find the value of capacitance,

  C = \frac{1}{4\pi ^{2} f^{2}  L }

  C = \frac{1}{4\times 9.85 \times (731 \times 10^{3} )^{2} \times 250 \times 10^{-3}  }

  C = 1.89 \times 10^{-13} F

Therefore, the value of variable capacitor is 1.89 \times 10^{-13} F

6 0
3 years ago
A 66 kg student stands on a very light, rigid board that rests on a bathroom scale at each end as shown in (Figure 1).
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The reading on the scale is equal to the weight of the student which is 646.8 N.

The given parameters;

  • <em>mass of the student, m = 66 kg</em>

The reading on the scale is determined by applying Newton's second law of motion as follows;

R = ma+ mg\\\\R = m(a + g)\\\\R = m(0 + g)\\\\R = mg

<em>The student is at rest and not accelerating, a =0</em>

R = 66 \times 9.8\\\\R = 646.8 \ N

Thus, the reading on the scale is equal to the weight of the student which is 646.8 N.

The complete question is below:

A 66 kg student stands on a very light, rigid board that rests on a bathroom scale at each end as shown in (Figure 1). what is the reading on the scale?

Learn more about Newton's second law here: brainly.com/question/3999427

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