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NikAS [45]
2 years ago
10

I can use everything on my body to advance the ball but what?

Physics
1 answer:
ddd [48]2 years ago
5 0
The answer would be hands! hope this helps
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What happens if the air pressure in your throat and outside body is less than the air pressure in your middle ear when you swall
andrezito [222]

Answer:

When the air pressure in the throat and outside the body is less than the air pressure in the middle ear, barotrauma occurs.

Explanation:

Ear barotrauma is a medical condition that describes discomfort in the ear which is caused by pressure differences in the inner and outer ear drum.

Usually, the air pressure in the middle ear is the same as the air pressure in the throat and outside the body.

When we swallow, the eustachian tube opens up and air flows out of and into the middle ear, this balances the pressure. But if the eustachian tube is blocked, the air pressure in the throat and outer body become different from the air pressure in the middle ear.

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3 years ago
Vector a cross b whole square<br><br>​
skad [1K]

Answer:

Kriss go Brrrrrrrrrrr

Explanation:

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3 years ago
What two factors are a part of thermohaline circulation
Margarita [4]

Answer:

These deep-ocean currents are driven by differences in the water's density, which is controlled by temperature (thermo) and salinity (haline). This process is known as thermohaline circulation.

Explanation:

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3 years ago
What is not a raquet sport
uysha [10]

Answer:

what

Explanation:

Racket sports include tennis, badminton, squash or any other sport where you use rackets to hit a ball or shuttlecock to play. They can be played competitively or just for fun and are a great form of physical activity.

7 0
3 years ago
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
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