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2 years ago

Hi! I really need help on these questions, it would be great if you can answer them!

Physics

2 answers:

Nikitich [7]2 years ago

6 0

1: Equilibrium

[] When two forces are equal in opposite directions, they are in a state of equilibrium

2: Friction

[] It can be hard to sometimes slide a paper across a table, roll a car across the ground, etc etc. This is because of a force called friction which resists motion

3: Displacement

[] Work is only "done" when the object or surroundings are having movement, or in motion, and this creates displacement

4: Force

[] A force is a push/pull in any direction, this is a simple definition so not much can be added.

5: Inertia

[] Inertia is a force that likes to remain unchanged, or without movement

6: Effort

[] "Meg is holding up weight without movement. She hasn't done any work since it did not move, but is tired. Why?" Effort! She still put in effort even though no "work" happened

7: Work

[] Force times displacement is the equation for work

8: Motion

[] Look back to #3, when there is motion we can tell work is done

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

Ilya [14]2 years ago

5 0

Equilibrium, friction, displacement, force, inertia

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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:

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