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ANTONII [103]
3 years ago
12

Which process is an example of a physical change?

Physics
1 answer:
svlad2 [7]3 years ago
6 0

Flattening is a example of a physical change.

Hope this helps! :D

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rusak2 [61]

Answer:

9.6

Explanation:

to convert km to miles multiply by 1.609

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How can scientists study the inner structure of an atom?
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A microscope is one way
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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
A car of mass 1,000 kilograms is moving initially at the speed of 22 meters/second. When the brakes are applied, it takes the ca
Mariulka [41]

i believe it is C....tell me if im right please<3

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3 years ago
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A small ball of charge Q and mass m has a velocity v at infinity. It collides head-on with a ball of the same charge and mass wh
Hatshy [7]

Answer:

Explanation:

Kinetic energy of ball in motion = 1/2 m v² . Potential energy = 0

Let the minimum distance between the balls be d  on collision.

Electric potential energy at that time= k Q²/d , Here kinetic energy is converted into potential energy . So

1/2 m v² = kQ²/d

d =2 k Q² / mv²,= 18 x 10⁹ x Q²/ m v².

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