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Alexxandr [17]
3 years ago
7

If a rock is dropped from the top of a tower at the front of it and takes 3.6 seconds to hit the ground. Calculate the final vel

ocity of the penny in m/s. What is the height of the tower?
Physics
1 answer:
expeople1 [14]3 years ago
6 0

Answer:

35.28m/s; 63.50m

Explanation:

<u>Given the following data;</u>

Time, t = 3.6 secs

Since it's a free fall, acceleration due to gravity = 9.8m/s²

Initial velocity, u = 0

To find the final velocity, we would use the first equation of motion;

V = u + at

Substituting into the equation, we have;

V = 0 + 9.8 * 3.6

V = 35.28m/s

Therefore, the final velocity of the penny is 35.28m/s.

To find the height, we would use the second equation of motion;

S = ut + \frac {1}{2}at^{2}

Substituting the values into the equation;

S = 0(3.6) + \frac {1}{2}*9.8*(3.6)^{2}

S = 0 + 4.9*12.86

S = 0.5 *36

S = 63.50m

Therefore, the height of the tower is 63.50m.

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When a certain gas under a pressure of 4.90 106 pa at 20.0°c is allowed to expand to 3.00 times its original volume, its final p
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As per question the initial states of the gases are given as

INITIAL STATE:                                          FINAL STATE:

p_{1} =4.90106 pa                       p_{2} =1.06106 pa

v_{1} =v[say]                                 v_{2} =3v[say]

T_{1} =20 degree celcius  =293 K              T_{2} =?

AS  per combined gas equation obtained from the combination of Boyle's law and Charles law [Basic ideal gas laws]

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Hence T_{2} =\frac{p_{2} v_{2}T_{1}  }{p_{1} v_{1} }

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                    =190.3 K [ANS]

                 

   

6 0
3 years ago
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