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crimeas [40]
3 years ago
8

A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a sp

eed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?
Physics
1 answer:
Aliun [14]3 years ago
5 0

Answer: velocity = 79.6m/s, distance covered = 139.6m and velocity when distance is 10.0m = 61.611m/s

Explanation:

First step is to take a direction of motion, so to avoid issues with negative sign, I take my direction of motion as the positive y axis which makes the parameters positive.

The object is of a free fall thus it has a constant acceleration of g = 9.8m/s².

a) initial velocity (u) = 60m/s, g = 9.8m/s², time taken (t) =2s, we need to get velocity at that time using the formulae below

v = u + at

v = 60 + (9.8)* 2

v = 60 + 19.6

v = 79.6m/s

b) to get the distance traveled at t =2s, we use the formulae below

s =ut + 1/2at² where s = distance covered.

s = 60 * 2 + 1/2 * 9.8 * 2²

s = 120 + 1/2 *9.8 * 4

s = 120 + 9.8 * 2

s = 120 + 19.6

s = 139.6m

c) to get the velocity (v) when distance traveled (s) = 10m, we use the formulae below.

v² = u² + 2as

v² = 60² + 2( 9.8)(10)

v² = 3600 + 196

v² = 3796

v = √3796

v = 61.611m

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C.Vacuum

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3 years ago
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3 years ago
A person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force F1 and the other hand holding
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Answer:

F1= 196 N

F2= 392 N  

Explanation:

Given:

length of the plank = 2 m

mass of the plank = 20 kg

Weight of the plank = 20 x 9.8 =196 N  

Torque due to the weight of the plank with respect to the pivoted end (i.e the end held by the hand) Counter clockwise torque  = 196 x cog of wood

= 196 x 1 = 196 Nm

Clockwise torque = F2 x 0.5

for the balanced case  

F2 x 0.5 = 196  

F2 = 196/ 0.5

F2= 392 N  

now,

the net force

Net downward force  =Net upward force

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4 0
3 years ago
At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 32.8105 cm on a spring with a sprin
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Answer:

0.175 m/s

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32.8105 cm = 0.328105 m

According to the law of conservation in energy, as the spring oscillates, its kinetic energy is converted to spring elastic energy and vice versa. So the maximum speed of the bananas occur when spring elastic energy is 0 and vice versa, spring elastic energy is maximum (at amplitude) with kinetic energy, and speed, is 0

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v^2 = \frac{13.9602*(0.328105)^2}{49.0943} = 0.0306

v = \sqrt{0.0306} = 0.175 m/s

3 0
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