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3 years ago

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A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in

the target holder and the steel ball she uses has a mass of 0.0120 kg. She finds that the target ball travels a distance of 1.50 m after it is struck. Assume g = 9.80 m/s2. What is the kinetic energy (in joules) of the target ball just after it is struck?

1 answer:

dexar [7]3 years ago

6 0

Answer:

K = 0.076 J

Explanation:

The height of the target, h = 0.860 m

The mass of the steel ball, m = 0.0120 kg

Distance moved, d = 1.50 m

We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.

$h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}$

Put all the values,

$t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s$

The velocity of the ball is :

$v=\dfrac{1.5}{0.418}\\\\= $$3.58\ m/s$

The kinetic energy of the ball is :

$K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J$

So, the required kinetic energy is 0.076 J.

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notka56 [123]

Answer:

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Given:

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Solution:

To calculate the fractional intensity, we use the given formula:

$\frac{I}{I_{max}} = \frac{sin^{2}\delta }{\frac({\delta}{2})^{2}}$ (1)

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For very small angle:

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$\theta = angle of deviation$

Using eqn (2):

$\delta = \frac{\pi \times 0.42\times 10^{-3}\times 4.1\times 10^{-3} }{546.1\times 10^{-9}\times 1.3} = 7.6202 radians$

Now, using eqn (1):

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Answer:

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Explanation:

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A girl wants to count the steps of a moving escalator which is going up. If she counts 60 steps going up and 90 coming down. How

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Leviafan [203]

Answer:

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Thank you for reading.

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