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aleksandrvk [35]
3 years ago
8

5. A 10 kg ball is traveling at the same speed as a 1 kg ball. Compared to the 10 kg ball, the 1 kg ball has (2 points)

Physics
1 answer:
suter [353]3 years ago
4 0
What’s the rest of the question or is that it?
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What will be the force if the particle's charge is tripled and the electric field strength is halved? Give your answer in terms
Alexus [3.1K]

Answer:

1.5F

Explanation:

Using

E= F/q

Where F= force

E= electric field

q=charge

F= Eq

So if qis tripled and E is halved we have

F= (E/2)3q

F= 1.5Eq=>> 1.5F

4 0
3 years ago
Technology is the application of science to make products that are useful to people. True or False
alekssr [168]
False false false false
5 0
3 years ago
Read 2 more answers
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
What’s the kinetic energy of the roller coaster at the top and bottom of the hill? Use . A kiddie roller coaster car has a mass
serious [3.7K]
K.E1=1/2×100×3²
=50×9
=450J
K.E2=1/2×100×36
=50×36
=1800J
0 0
4 years ago
Read 2 more answers
1. How is a whistling teapot like a volcano?
oksano4ka [1.4K]
Because of the build up of pressure. There is so much steam coming from such a compressed point, it’s coming out in force.

Now think of that same spot being closed, it only has one place to go but it can’t leave, so that pressure will build and build and then BOOM, it explodes.

In short, the answer is the pressure being released from a small point, and how that energy is released.
8 0
3 years ago
Read 2 more answers
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