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4 years ago

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A 0.0450-kg golf ball initially at rest is given a speed of 25.0 m/s when a club strikes. If the club and ball are in contact fo

r 2.00 ms, what average force acts on the ball? *

1 answer:

LuckyWell [14K]4 years ago

5 0

Answer:

Average force, F = 562.5 N

Explanation:

Mass of the golf ball, m = 0.045 kg

Initially, it is at rest, u = 0

Final speed of the ball, v = 25 m/s

The club and the ball are in contact for, $t=2\ ms=2\times 10^{-3}\ s$

We need to find the average force acting on the ball. It can be calculated using the formula as :

$F\times t=m(v-u)$

$F=m\dfrac{v-u}{t}$

$F=0.045\times \dfrac{25}{2\times 10^{-3}}$

F = 562.5 N

So, the average force acting on the ball is 562.5 N. Hence, this is the required solution.

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Explanation:

The question is incomplete: however, we can still answer as follows.

The mass of a radioactive sample after a time t is given by the equation:

$m(t)=m_0 (\frac{1}{2})^{\frac{t}{\tau}}$

where:

$m_0$ is the mass of the radioactive sample at t = 0

$\tau$ is the half-life of the sample

This means that the mass of the sample halves after one half-life.

We can rewrite the equation as

$\frac{m(t)}{m_0}=(\frac{1}{2})^{\frac{t}{\tau}}$

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Therefore, after t = 1 days, the fraction of radioisotope left in the body is

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A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

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For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

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PE: is the potential energy = mgh

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$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

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$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

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$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The <em>kinetic energy</em> of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

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