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Gemiola [76]
3 years ago
13

Explain why, in terms of forces, there is a risk of head injury when diving from this height. Suggest why the high divers would

choose to enter the water feet first.
Physics
1 answer:
podryga [215]3 years ago
4 0

Answer:

Due to lower risk of injury or damage.

Explanation:

The high divers would choose to enter the water from the feet first because there is low risk of injury. The brain is the most important part of the body which very sensitive to any small injury. Small injury to brain leads to big problems in life. High divers can reach speeds of nearly 60 mph and enters about 28m into the water in about three seconds which can damage the head region if comes in contact with the ground so this is the reason the high divers avoid of entering in the water through their heads and choose entering through their feet.

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What is the acceleration of this object? The object's mass is 60 kg.
slamgirl [31]
We know, F = m*a
Here, F = 600-400 = 200 N downwards 
m = 60 Kg

Substitute their values, 
a = 200/60
a = 20/6
a = 10/3
a = 3.33 m/s² downwards

In short, Your Answer would be 3.33 m/s² downwards

Hope this helps!
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if a person can jump maximum along distance of 3m ,on the earth how far could be jump on the moon where acceleration due to grav
allochka39001 [22]

Answer:

The person can jump 48 m on the Moon

Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, R_{max} = 3 m

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The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, R_{max}, of a projectile motion is found by using the following formula

R_{max} = \dfrac{u^2}{g}

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore, we have;

R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, g_{Moon} = 1/16 × g

∴ g_{Moon} = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

R_{max \ Moon} = \dfrac{u^2}{g_{Moon}}   = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is R_{max \ Moon} ≈ 48 m

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