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3 years ago

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If a rock climber accidentally drops a 77.5-g piton from a height of 215 meters, what would its speed be just before striking th

e ground? ignore the effects of air resistance.

1 answer:

Viktor [21]3 years ago

4 0

Let us list out what we know from the question.

Initial Velocity $V_{i} = 0$ since the piton is 'dropped'.

Vertical Displacement of the piton D = 215 m

Acceleration due to gravity $a = 9.8 m/s^{2}$

Final Velocity $V_{f} = ?$

Using the equation, $V^{2} _{f} = V^{2} _{i} + 2aD$ and plugging in the known values, we get

$V^{2} _{f} = 0^{2} + 2(9.8)(215)$

Simplifying by taking square-root on both sides gives us $V_{f} = 64.915 m/s$

Thus, the speed of the piton just before striking the ground is 65 m/s.

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Nitrogen at 100 kPa and 25oC in a rigid vessel is heated until its pressure is 300 kPa. Calculate (a) the work done and (b) the

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Answer:

A. The work done during the process is W = 0

B. The value of heat transfer during the process Q = 442.83 $\frac{KJ}{kg}$

Explanation:

Given Data

Initial pressure $P_{1}$ = 100 k pa

Initial temperature $T_{1}$ = 25 degree Celsius = 298 Kelvin

Final pressure $P_{2}$ = 300 k pa

Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.

⇒ $V_{1}$ = $V_{2}$ ------------- (1)

Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.

⇒ P ∝ T

⇒ $\frac{P_{2} }{P_{1}}$ = $\frac{T_{2} }{T_{1}}$

⇒ Put all the values in the above formula we get the final temperature

⇒ $T_{2}$ = $\frac{300}{100}$ × 298

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From equation (1), $V_{1}$ = $V_{2}$ so work done W = P × 0 = 0

⇒ W = 0

Therefore the work done during the process is zero.

Heat transfer during the process is given by the formula Q = m $C_{v}$ ( $T_{2}$ - $T_{1}$ )

Where m = mass of the gas = 1 kg

$C_{v}$ = specific heat at constant volume of nitrogen = 0.743 $\frac{KJ}{kg k}$

Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )

⇒ Q = 442.83 $\frac{KJ}{kg}$

Therefore the value of heat transfer during the process Q = 442.83 $\frac{KJ}{kg}$

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Two +1 C charges are separated by 30000 m, what is the magnitude of<br> the force?

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Answer:

<em>The magnitude of the force is 10 N</em>

Explanation:

<u>Coulomb's Law</u>

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Written as a formula:

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d= The distance between the objects

We have two identical charges of q1=q2=1 c separated by d=30000 m, thus the magnitude of the force is:

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$\displaystyle F=9\cdot 10^9\frac{1*1}{30000^2}$

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