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Irina-Kira [14]
3 years ago
6

The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportiona

l to the square of the distance from the source. If two light sources, one four times as strong as the other, are placed 14 ft apart, how far away from the stronger light source should an object be placed on the line between the two sources so as to receive the least illumination? (Round your answer to two decimal places.)
Physics
1 answer:
34kurt3 years ago
7 0

Answer:

d=4.67ft from the weaker source

Explanation:

The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source:

Illumination=k*S/d^{2}   first source

Illumination=k*4S/(14-d)^{2}   second source (stronger)

If the illumination is the same for the two sources:

k*S/d^{2}=k*(4S)/(14-d)^{2}

We solve to find d:

2d=14-d

d=14/3=4.67ft from the weaker source

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1. The bowling ball have more potential energy as it sit on top of the building

2.The bowling ball have same potential energy and kinetic energy as it is half way through its fall

3. The bowling ball have more kinetic energy just before it hits the ground

4. The potential energy of the bowling ball as it sits on top of the building is 784J

5. The potential energy of the ball as it is half way through the fall, 20 meters high is 392J

6. The kinetic energy of the ball as it is half way through the fall is 392J

7. The kinetic energy of the ball just before it hits the ground is 784J

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Calculating potential energy and kinetic energy for all the instances,

1. ball on top of a 40 meters tall building

Potential energy at the top of building with a height of 40m = mgh

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At the top pf the building since v=0 kinetic energy is zero

2. half way through a fall off a building that is 40 meters tall and travelling 19.8 meters per second

Potential energy when it is half way through fall = mgH

where H represents new height that is equal to 20m

hence P.E=mgH=2*9.8*20= 392J

Kinetic energy  of the ball is \frac{1}{2} mv^{2}  = \frac{1}{2} *2*19.8^{2}=392.04J

3.  Just about to hit the ground from a fall off a building that is 40 meters tall and travelling 28 meters per second.

The potential energy of ball just before it hits the ground = mgh= 2*9.8*0=0J

kinetic energy =\frac{1}{2} mv^{2}= 784J

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