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Colt1911 [192]
3 years ago
9

In the child's game of tetherball, a rope attached to the top of a tall pole is tied to a ball. Players hit the ball in opposite

directions in an attempt to wrap the ball and rope around the pole. Assume the rope has negligible mass and that resistive forces, such as air resistance and friction, can be neglected. As the ball wraps around the pole between hits, how does the angular speed of the ball change? a.it decreases b. it stays the same c.it increases
Physics
1 answer:
Eddi Din [679]3 years ago
7 0

Answer:

option C

Explanation:

the ball is moving circular around the pole

Angular momentum of the system is constant

              J = I ω

now,

               \omega\ \alpha\ \dfrac{1}{I}

               \omega\ \alpha\ \dfrac{1}{mr^2}

               \omega\ \alpha\ \dfrac{1}{r^2}

the rope radius is decreasing as it revolving around the pole

angular speed is inversely proportional to radius.

so, the angular speed will increase.

The correct answer is option C

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During the time interval from 0.0 to 10.0 s, the position vector of a car on a road is given by x(t) = a + bt + ct2, with a = 17
Juli2301 [7.4K]

The car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

<h3>Average velocity of the car</h3>

The average velocity of the car is calculated as follows;

x(t) = a + bt + ct2

v = dx/dt

v(t) = b + 2ct

v(0) = -10.1 m/s + 2(1.1)(0) = -10.1 m/s

v(10) = -10.1 + 2(1.1)(10) = 11.9 m/s

<h3>Average velocity</h3>

V = ¹/₂[v(0) + v(10)]

V = ¹/₂ (-10.1  + 11.9 )

V = 0.9 m/s

Thus, the car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

Learn more about velocity here: brainly.com/question/4931057

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1 year ago
Which type of energy can be sensed by the eyes?
gayaneshka [121]

Answer:

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7 0
3 years ago
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Shanti is riding on a train that is moving at a speed of 90 km/h. He is carrying a power cord for his phone that is 1.2 m long.
faltersainse [42]

Answer:equal to 1.2 m

Explanation: i took the test

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The work done by an external force to move a -8.50 μC charge from point a to point b is 6.10×10−4 J . If the charge was started
bekas [8.4K]

Answer:

-54.12 V

Explanation:

The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.

W=\Delta E\\W=\Delta K+\Delta U\\W=K_f+\Delta U\\\Delta U=W-K_f\\\Delta U=6.10*10^{-4}J-1.50*10^{-4}J\\\Delta U=4.60*10^{-4}J

The change in electrostatic potential energy \Delta U, of one point charge q is defined as the product of the charge and the potential difference.

\Delta U=qV\\V=\frac{\Delta U}{q}\\V=\frac{4.60*10^{-4}J}{-8.50*10^{-6}C}\\V=-54.12 V

5 0
3 years ago
Two identical charged pith balls are brought together to touch each other. They are then
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Answer:

-17.5 nC

Explanation:

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answer is -17.5 nC

4 0
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