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muminat
3 years ago
14

A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the grou

nd exerts a +55 N force on the ball. What is the magnitude of the velocity after it hits the ground?
Physics
2 answers:
FinnZ [79.3K]3 years ago
7 0

<u>Answer</u>

= 9.25 m/s


<u>Explanation</u>

The Newton's second law of motion states that, the change in momentum is directly propotional to the force producing it and it takes place in the direction of force.

F = ma

f = m(v-u)/t

ft = m(v-u)

∴ 55 × 45/1000 = 0.060(v - -32)

2.475 = 0.06(v + 32)

2.475/0.6 = v + 32

41.25 = v + 32

v = 41.25 -32

= 9.25 m/s

Tanzania [10]3 years ago
4 0

On E2020 the answer is 9.3 m/s

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Answer:

C is halved

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f is the frequency

\lambda is the wavelength

From the equation above, we see that for a given wave, if the wave is travelling in the same medium (and so, its speed is not changing), then the frequency and the wavelength are inversely proportional to each other.

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2 years ago
A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th
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The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = 9.8\ m/s^2

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

                Kinetic energy K.E = \frac{1}{2}mv^2.

From above we know that -

   Kinetic energy at the bottom of the cliff = potential energy at a height h

                 \frac{1}{2}mv^2=\ mgh

                ⇒ v^2=\ 2gh

                ⇒ h=\ \frac{v^2}{2g}

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3 years ago
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