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Lina20 [59]
3 years ago
9

Select the correct answer. What happens to the sum of the ball’s kinetic energy and potential energy as the ball rolls from poin

t A to point E? Assume there’s no friction between the ball and the ground. A. The sum decreases. B. The sum increases. C. The sum remains the same. D. The sum always equals zero.
Physics
2 answers:
barxatty [35]3 years ago
7 0

Answer:

It might be the sum is the same

Explanation:

Correct me if i am wrong but there is no friction so that means there would not be anything slowing down the ball, therefore the ball wouldn't change or the energy wouldn't change because the ball would never stop.

AlekseyPX3 years ago
3 0

The sum of the ball’s kinetic energy and potential energy remains the same as the ball rolls from point A to point E ( if there’s no friction between the ball and the ground).

Answer: Option C

<u>Explanation: </u>

Based on the law of mechanical energy's conservation, the sum of both kinetic and potential energies i.e. the total amount of mechanical energy remain conserved even in the absence of dissipative forces (friction or air resistance) in a bound system.

                    E = Kinetic + Potential

The kinetic energy is high when ball goes downward but potential energy decreases and reverse happens when ball goes up. But in these case, the sum energy would be constant one.

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What is the amount of thermal energy needed to make 5 kg of ice at - 10 °C to
agasfer [191]

Answer:

The amount of thermal energy needed is 15167500 joules.

Explanation:

By First Law of Thermodynamics, we see that amount of thermal energy (Q), in joules, is equal to the change in internal energy. From statement we understand that change in internal energy consisting in two latent components (U_{l,ice}, U_{l,steam}), in joules, and two sensible component (U_{s,w}), in joules, that is:

Q = U_{l,ice} + U_{s, w} + U_{s,ice} + U_{l,steam} (1)

By definitions of Sensible and Latent Heat, we expanded the formula:

Q = m\cdot (h_{f,w}+h_{v,w}+c_{ice}\cdot \Delta T_{ice}+c_{w}\cdot \Delta T_{w}) (2)

Where:

m - Mass, in kilograms.

h_{f,w} - Latent heat of fussion of water, in joules per kilogram.

h_{v,w} - Latent heat of vaporization of water, in joules per kilogram.

c_{ice} - Specific heat of ice, in joules per kilogram per degree Celsius.

c_{w} - Specific heat of water, in joules per kilogram per degree Celsius.

\Delta T_{ice} - Change in temperature of ice, measured in degrees Celsius.

\Delta T_{w} - Change in temperature of water, measured in degrees Celsius.

If we know that m = 5\,kg, h_{f,w} = 3.34\times 10^{5}\,\frac{J}{kg}, h_{v,w} = 2.26\times 10^{6}\,\frac{J}{kg}, c_{ice} = 2.090\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}, c_{w} = 4.186\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}, \Delta T_{ice} = 10\,^{\circ}C and \Delta T_{w} = 100\,^{\circ}C, then the amount of thermal energy is:

Q = 15167500\,J

The amount of thermal energy needed is 15167500 joules.

7 0
3 years ago
A 20-cm long solenoid consists of 100 turns of a coil of radius r = 3.0 cm. A current of Io in the coiled wire produces a magnet
Romashka-Z-Leto [24]

Answer:

vi) Double the current in the wire, and double the number of turns in the 20-cm long solenoid

Explanation:

The magnetic field inside the solenoid and the current flowing in the coil of solenoid are related to each other by the following equation

B₀=μ₀nI₀

Where,

B₀ is the magnetic field in the middle of solenoid

n is the number of turns in the coil of solenoid

I₀ is the current flowing in the coil of solenoid

In the above equation, as μ₀ is a constant so the magnetic field will be directly proportional to the number of turns multiplied by the current. So, changing the radius of the coil or length of the coil will have no effect on the magnetic field.

As we have to increase the magnetic field by 4 times, we need to double the current as well as the number of turns as mentioned in the option vi.

3 0
3 years ago
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Answer:

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3 years ago
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Answer:

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D .F= G (4mM)/(2r)² = G (mM)/r²

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