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fredd [130]
3 years ago
7

Question 14 (1 point)

Physics
1 answer:
mash [69]3 years ago
7 0
(c) as the change in the dependent variable is in direct CORRELATION to the change in the independent variable.
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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
n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at
Volgvan

Answer:

a)-1.014x 10^{-7J

b)3.296 x  10^{-7J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

xb=3.4m

a)the gravitational potential energy is given by

U_{i = -GMaMb/ d

U_{i= - 6.67 x 10^{-11} x 47 x 110/ 3.4 => -1.014x 10^{-7J

b) at d= 0.8m (3.4-2.6) and U_{i=-1.014x 10^{-7J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

K_{i + U_{i= K_{f + U_{f

As sphere starts from rest and sphere A is fixed at its place, therefore K_{i is zero

U_{i= K_{f + U_{f

The final potential energy is

U_{f= - GMaMb/d

Solving for 'K_{f '

K_{f = U_{i + GMaMb/d => -1.014x 10^{-7 + 6.67 x 10^{-11} x 47 x 110/ 0.8

K_{f = 3.296 x  10^{-7J

6 0
3 years ago
Bond formed when one or more electrons are transferred from one atom to another. True or False
Finger [1]

Answer: true

Explanation:

8 0
3 years ago
Starting from rest, a 2.1x10^-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exert
Rainbow [258]

Answer:

a) The flea's speed when it leaves the ground is v=1.88m/s

b) The flea move 18cm upward while it is pushing off

Explanation:

Hi

<u>Knwons</u>

Mass m=2.1\times 10^{-4} kg, Work W=3.7\times 10^{-4} J and Force F=0.41N

a) Here we are going to use W=\frac{1}{2} mv^{2}, so v=\sqrt{\frac{2W}{m} }= \sqrt{\frac{2(3.7\times 10^{-4} J)}{2.1\times 10^{-4} kg} }=1.88m/s

a) Here we are going to use W=mgh, so h=\frac{W}{mg}= \frac{3.7\times 10^{-4} J}{(2.1\times 10^{-4} kg)(9.8m/s^{2})} =0,1797m or 18cm approx.

7 0
3 years ago
Se o Universo tem uma idade definida, de aproximadamente 13,7 bilhões de anos, o que existiu antes do big-bang?
UkoKoshka [18]
Ninguém sabe exatamente
4 0
3 years ago
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