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Harlamova29_29 [7]
2 years ago
14

Electronic flash units for camera contain a capacitor for storing the energy used to produce the flash. In one such unit, the fl

ash lasts for 1675s with an average light power output of 2.70 x 105 W . (a) If the conversion of electric energy to light is 95% efficient ( the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash
Physics
1 answer:
nadezda [96]2 years ago
8 0

Answer:

420J

Explanation:

Power is the time rate of change in energy. Power is the ratio of energy to time. The S.I unit of power is in watts.

Given that the flash lasts for 1/675 s, power output is 2.7 * 10⁵ W. Hence:

Power = Energy / time

Substituting:

2.7 * 10⁵ W = Energy / (1/675)

Energy = 2.7 * 10⁵ W * 1/675 = 400J

Therefore the energy emitted as light is 400J.

Since the conversion of electric energy to light is 95% efficient, hence the energy stored as electrical energy is:

Energy(capacitor) = 5% of 400J + 400J = 0.05*400 + 400

Energy(capacitor) = 420J

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A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s
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A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is:   w_{c} =m_{c} g   and maximum tension is T=2.50 m_{c} g=1.41*10^4N

total mass and weight is :

M =m_{c}+ m_{b} =740kg+550kg=1290 kg

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T-M_{g} =Ma_{y}

a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg

=0.645m/s^2

B)

maximum acceleration

a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0

using y-y_{0} =v_{oy} t+1/2(a_{y} )t^2

to solve for t

t=\sqrt{2(y-y_{0} )/a_{y} }

t=\sqrt{2(119m)/0.645m/s^2} =19.20s

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3 years ago
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I is the moment of inertia

m is the mass

r is the distance between the arbitrary point and the point mass

The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.

The total moment of inertia of the system is the sum of the moments of each mass, i.e.

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