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Oduvanchick [21]
3 years ago
12

A car travels 20 km southwards and then travel another 20km westward what is the displacement from the rest position

Physics
1 answer:
Juli2301 [7.4K]3 years ago
7 0

\text{Hello there!}\\\\\text{We're trying to find displacement}\\\\\text{We would use the Pythagorean theorem to find the displacement:}\\a^2+b^2=c^2\\\\\text{We know that the car goes 20 km south and 20 km west}\\\\\text{When you draw it out, it should look like a triangle, you would be finding}\\\text{the length of the hypotenuse}\\\\\text{Plug in 20 to} \,\,a^2\,\, \text{and}\,\, b^2}\\\\20^2+20^2=c^2\\\\\text{Solve:}\\\\20^2+20^2=c^2\\\\400+400=c^2\\\\800=c^2\\\\\sqrt800=\sqrt c\\

\large\boxed{28.28}\\\\\normalsize\text{The displacment would be 28.28}

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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
3 years ago
A cannon fires a shell straight upward; 2.3 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming
PtichkaEL [24]

Answer:

The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.                    

Explanation:

Let u is the initial speed of the launch. Using first equation of motion as :

u=v-at

a=-g

u=v+gt\\\\u=17+9.8\times 2.3\\\\u=39.54\ m/s

The velocity of the shell at launch and 5.4 s after the launch is given by :

v=u-gt\\\\v=39.54-9.8\times 5.4\\\\v=-13.38\ m/s

So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

6 0
3 years ago
A toy car pushed across the floor is observed to slow down and stop based on Newton’s first law of motion what is the best concl
Maksim231197 [3]

An object in motion eventually stops because of friction.

<h3>What is Newton’s first law of motion?</h3>

Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or continue its moving in a straight line at constant speed unless force is applied on it.

So we can conclude that An object in motion eventually stops is the correct answer.

Learn more about law here: brainly.com/question/820417

3 0
2 years ago
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