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Wewaii [24]
3 years ago
11

Nicole throws a ball straight up. Chad watches the ball from a window 6.10m above the point where Nicole released it. The ball p

asses Chad on the way up, and it has a speed of 12.0m/s as it passes him on the way back down.How fast did Nicole throw the ball?
Physics
1 answer:
solong [7]3 years ago
3 0

Answer:

v=16.234m/s

Explanation:

Given data

Distance S=6.10 m

Initial Speed u=12.0 m/s

To find

Final Speed v

Solution

From third equation of motion we know that:

v^{2}-u^{2}=2aS

where

v is final velocity

u is initial velocity

a is acceleration

S is distance

Substitute the given values to find v

v^{2}=u^{2}+2aS\\v^{2}=(12.0m/s)^{2} +2(9.8m/s^{2} )(6.10m)\\v^{2}=263.56\\  v=\sqrt{263.56}\\ v=16.234m/s

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3 years ago
As a system expands, it absorbs 52.5 J of energy in the form of heat from the surroundings. The piston is working against a pres
Kaylis [27]

Answer:

Vi = 0.055 m³ = 55 L

Explanation:

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ΔQ = ΔU + W

where,

ΔQ = Heat absorbed by the system = 52.5 J

ΔU = Change in Internal Energy = -102.5 J (negative sign shows decrease in internal energy of the system)

W = Work Done in Expansion by the system = ?

Therefore,

52.5 J = - 102.5 J + W

W = 52.5 J + 102.5 J

W = 155 J

Now, the work done in a constant pressure condition is given by:

W = PΔV

W = P(Vf - Vi)

where,

P = Constant Pressure = (0.5 atm)(101325 Pa/1 atm) = 50662.5 Pa

Vf = Final Volume of System = (58 L)(0.001 m³/1 L) = 0.058 m³

Vi = Initial Volume of System = ?

Therefore,

155 J = (50662.5 Pa)(0.058 m³ - Vi)

Vi = 0.058 m³ - 155 J/50662.5 Pa

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7 0
4 years ago
In a thundercloud there may be an electric charge of +40 C near the top of the cloud and -40 C near the bottom of the cloud. The
Ratling [72]

Answer:

Electric force, F=-3.59\times 10^6\ N

Explanation:

Given that,

Electric charge 1, q_1=+40\ C

Electric charge 2, q_2=-40\ C

Distance, d=2\ km=2\times 10^3\ m

To find,

The electric force between these two sets of charges.

Solution,

There exists a force between two electric charges and this force is called electrostatic force. It is equal to the product of electric charged divided by square of distance between them.

F=k\dfrac{q_1q_2}{d^2}

k is the electrostatic constant

F=8.99\times 10^9\times \dfrac{40\times (-40)}{(2\times 10^3)^2}

F=-3.59\times 10^6\ N

So, the electric force between these two sets of charges is -3.59\times 10^6\ N.

5 0
3 years ago
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