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bekas [8.4K]
3 years ago
11

Suppose you toss a tennis ball upward.a) Does the kinetic energy of the ball increase or decrease as it moves higher?b) What hap

pens to the potential energy of the ball as it moves higher?c) If the same amount of energy were imparted to a ball the same size as a tennis ball but of twicethe mass, how high would it go in comparison to the tennis ball?
Physics
1 answer:
Lilit [14]3 years ago
7 0

Answer:

a) No. The kinetic energy of the ball decreases.

b) The potential energy of the ball increases.

c) The ball would go half of the original distance.

Explanation:

a) The kinetic energy would be converted to potential energy as the ball goes higher. Since the total mechanical energy is conserved, the kinetic energy would decrease.

b) The potential energy of the ball would increase. Since the total mechanical energy of the ball is conserved, the ball would lose speed, and therefore kinetic energy. In order to compensate the loss of kinetic energy, the ball would gain potential energy as it goes higher.

c) The relation of the energy and mass is as follows:

K = \frac{1}{2}mv^2\\U = mgh

According to the energy conservation

K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv^2 + 0 = 0 + mgh\\\frac{1}{2}v^2 = gh

The maximum height that the ball reaches is proportional to the initial velocity. If the ball would be imparted with the same amount of energy, its final potential energy would be the same. However, in order to have the same potential energy (mgh), its height would be half of the original case.

mgh = (2m)g(\frac{h}{2})

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AveGali [126]

The best and most correct answer among the choices provided by your question is the fourth option or letter D. Trade winds blow towards the equator because t<span>he Equator receives the most heat energy.

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3 0
3 years ago
If the area of an iron rod is 10 cm by 0.5 cm and length is 35 cm. Find the value of resistance, if 11x10^-8 ohm.m be the resist
malfutka [58]

Answer:

Resistance of the iron rod, R = 0.000077 ohms    

Explanation:

It is given that,

Area of iron rod, A=10\ cm\times 0.5\ cm=5\ cm^2 = 0.0005\ m^2

Length of the rod, L = 35 cm = 0.35 m

Resistivity of Iron, \rho=11\times 10^{-8}\ \Omega-m

We need to find the resistance of the iron rod. It is given by :

R=\rho\dfrac{L}{A}

R=11\times 10^{-8}\times \dfrac{0.35\ m}{0.0005\ m^2}

R=0.000077 \Omega

So, the resistance of the rod is 0.000077 ohms. Hence, this is the required solution.

6 0
3 years ago
A bowling ball has a mass of 7.2 kg and a weight of 70.6 N. It moves down the bowling alley at 1 m/s and strikes a pin with a fo
Inga [223]
The answer is 15.0N its explained in newtons third law hope this helps:)

4 0
3 years ago
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A force of 20N pushes an object of mass 5.0kg along a rough surface of 5.0N​
Alika [10]

Answer:

I'm sorry, I don't think there is any answer to give seeing as no question has been asked

7 0
3 years ago
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A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotating t
koban [17]

Answer:

The coefficient of static friction between the object and the disk is 0.087.

Explanation:

According to the statement, the object on the disk experiments a centrifugal force due to static friction. From 2nd Newton's Law, we can represent the object by the following formula:

\Sigma F_{r} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R} (1)

\Sigma F_{y} = N - m\cdot g = 0 (2)

Where:

N - Normal force from the ground on the object, measured in newtons.

m - Mass of the object, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v - Linear speed of rotation of the disk, measured in meters per second.

R - Distance of the object from the center of the disk, measured in meters.

By applying (2) on (1), we obtain the following formula:

\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}

\mu_{s} = \frac{v^{2}}{g\cdot R}

If we know that v = 0.8\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}} and R = 0.75\,m, then the coefficient of static friction between the object and the disk is:

\mu_{s} = \frac{\left(0.8\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.75\,m)}

\mu_{s} = 0.087

The coefficient of static friction between the object and the disk is 0.087.

5 0
3 years ago
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