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Xelga [282]
3 years ago
9

If a car is traveling 103 miles per hour, how many miles and how much time would it take for another car to catch up to them fro

m a complete stop?
Physics
1 answer:
pentagon [3]3 years ago
3 0

Answer:

Required distance is 103 miles and the required time is 1 hour

Explanation:

Given;

speed of the car, v = 103 miles per hour

Speed is the given as the ratio of distance traveled to time taken for the motion.

The distance it will take another car to catch up to them from a complete stop is 103 miles and the time it will take the car is 1 hour.

Therefore, required distance is 103 miles and the required time is 1 hour.

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A man has 887.5 J of kinetic energy while running with a velocity of 5 m/s. What is his mass?
monitta

Answer:

The mass of the man is 71 kg

Explanation:

Given;

kinetic energy of the man, K.E = 887.5 J

velocity of the man, v = 5 m/s

The mass of the man is calculated as follows;

K.E = ¹/₂mv²

where;

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1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s
siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

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The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

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