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faust18 [17]
3 years ago
15

If i hold a glass full of air upside down, and i take it to the bottom of the swimming pool without tipping it, the density (thi

ckness) of the air inside the glass will be ________ at the surface.
Physics
1 answer:
s2008m [1.1K]3 years ago
6 0
The answer to fill in the blank in the question above is "greater than" based on the physic of the air density. The density of air is affected by the temperature and the pressure based on the ideal gas law. A high pressure will make the air becomes denser and the bottom of swimming pool has a higher pressure than the surface<span>.</span>
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A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15
Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

T-mg = m\frac{v^2}{R}

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

m \frac{v^2}{R} is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

T=m\frac{v^2}{R}

And by substituting the numerical values, we find

T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

T+mg=m\frac{v^2}{R}

And substituting the numerical values and re-arranging it, we find

T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

T=0\\mg = m\frac{v^2}{R}

and re-arranging the equation, we find

v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s

7 0
3 years ago
According to newton's first law, what is required to make an object slow down?
Luden [163]
I believe it is friction

3 0
3 years ago
Read 2 more answers
A(n) 30 kg boy rides a roller coaster. The acceleration of gravity is 9.8 m/s 2 . With what force does he press against the seat
forsale [732]

Answer:

Force, F = 187.42 N

Explanation:

It is given that,

Mass of boy, m = 30 kg

Acceleration due to gravity, a=9.8\ m/s^2

Radius of curvature of the roller coaster, r = 15 m

Speed of the car, v = 7.3 m/s

The force acting on the boy are force of gravity and the centripetal force. The net force acting on him is as follows :

F=mg-\dfrac{mv^2}{r}

F=m(g-\dfrac{v^2}{r})

F=30\times (9.8-\dfrac{(7.3)^2}{15})

F = 187.42 N

So, he press against the seat with a force is 187.42 N. Hence, this is the required solution.

5 0
3 years ago
HEY CAN ANYONE PLS ANSWER DIS!!!!!!!!!!!!
Salsk061 [2.6K]

Answer:

a is falsa

Explanation:

8 0
3 years ago
Mars has two moons. What does the orbital speed of these moons depend on?
Radda [10]

Answer:

their masses and their distances from Mars

Explanation:

Hope this helped

Also can u pls mark me brainliest

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