Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,

Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?

a = -1.47 m/
(a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/
(since same friction force is applied)

s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
The intensity on a screen 20 ft from the light will be 0.125-foot candles.
<h3>What is the distance?</h3>
Distance is a numerical representation of the length between two objects or locations.
The intensity I of light varies inversely as the square of the distance D from the source;
I∝(1/D²)
The ratio of the intensity of the two cases;

Hence, the intensity on a screen 20 ft from the light will be 0.125 foot-candles
To learn more about the distance refer to the link;
brainly.com/question/26711747
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