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Ann [662]
3 years ago
13

A sled slides down a hill, reaches the level surface, and eventually comes to a stop. Which fact ultimately explains why the sle

d stops?
A. The presence of an unbalanced force causes an object to move forward at constant velocity
B. The presence of an unbalanced force (e.g. friction) causes a moving object to stop
C. The presence of inertia causes a moving object to stop
D. The natural tendency of any object is to ultimately stop
Physics
1 answer:
Kitty [74]3 years ago
7 0

Answer:

B. The presence of an unbalanced force(e.g friction) causes a moving object to stop.

Explanation:

As the friction is that force that can stop the sled upon reaching the levelled surface so the option b is correct.

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The answer is "15.38\%"

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Background= 13 \ mv\\\\

corrected signal= 91 \ mv-13\ mv= 78\ mv\\\\

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\to \frac{p_t}{p_i}=\frac{12}{78}\times 100= 15.38\%

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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

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

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

4 0
2 years ago
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