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Elena L [17]
3 years ago
15

Write a question about how changing temperature affects the gas inside the sports ball

Physics
1 answer:
dmitriy555 [2]3 years ago
6 0

Answer:

What happens to the pressure of the air inside a sport ball when it is heated?

Explanation:

We can answer this question by thinking at what happen at microscopic level.

In fact, when the gas inside the ball is heated, the molecules of the gas start moving faster. As a result, the rate of collision of the molecules against the internal surface of the ball increases: and therefore, the pressure of the gas inside the ball increases.

We can also see this by looking at the ideal gas law, which states that:

pV=nRT

where

p is the gas pressure

V is the gas volume

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

In this situation, the volume of the gas V is constant (since the ball has a constant volume), the number of moles n is also constant, as well as R. So we can rewrite this as

p \propto T

so we see that the pressure is directly proportional to the temperature: therefore, when the ball is heated, the pressure inside the ball increases.

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a vector is 253 m long and points in a 55.8 degree direction, what’s the y and x- component of the vector?
BartSMP [9]

Well we know the hypotenuse of the triangle which is 253 m. And we know the angle of the triangle which is 55.8 degrees. So we want to find y. And to find y we use sin. And sin is a ratio, the ratio of the opposite leg, and hypotenuse. So sin(55.8) = y/253. Now we solve for y by multiplying both sides by 253. And finally we get 209.25 as the length of the y component.

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Cu ce viteza s-a deplasat un vehicul daca la ora 16:45 se afla in dreptul bornei kilometrice 169 iar la ora 17:50 se afla la bor
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3 0
3 years ago
a parent swings a 18.5 kg child in a circle of radius 1.05m, making 5 revolutions in 13.4s. what centripetal acceleration does t
Zolol [24]

Answer: 0.146 m/s^{2}

Explanation:

The <u>centripetal acceleration</u> a_{c} of an object moving in a uniform circular path is given by the following equation:

a_{c}=\frac{V^{2}}{r}  (1)

Where:

V is the tangential velocity

r=1.05 m is the radius of the circle

On the other hand, the tangential velocity  is expressed as:

V=\omega r (2)

Where \omega is the angular velocity, which can be found knowing the child makes 5 revolutions in 13.4s:

\omega=\frac{5 rev}{13.4 s}=0.37 rev/s (3)

Substituting (3) in (2):

V=(0.37 rev/s)(1.05 m) (4)

V=0.39 m/s (5)

Substituting (5) in (1):

a_{c}=\frac{(0.39 m/s)^{2}}{1.05 m}  (6)

Finally:

a_{c}=0.146 m/s^{2}  

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