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svetlana [45]
2 years ago
12

A charge of 0.14 C is moved from a position where the electric potential is 20 V to a position where the electric potential is 5

0 V. What is the change in potential energy of the charge associated with the change in position?
Physics
1 answer:
alukav5142 [94]2 years ago
7 0

The change in electric potential energy is given by:

ΔU = ΔVq

ΔU = change in PE, ΔV = potential difference, q = charge

Given values:

ΔV = 50V - 20V = 30V, q = 0.14C

Plug in and solve for ΔU:

ΔU = 30(0.14)

ΔU = 4.2J

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2. Do you think the density of the ice affected the melting rate of the ice, or do you think adding the objects affected the mel
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The density of ice does not affect its melting rate. Adding objects will affect the melting rate.

  • A physical process called melting or fusing causes a substance to change its phase from a solid to a liquid. This happens when the solid's internal energy rises, usually as a result of heat or pressure being applied, which raises the substance's temperature to the melting point.
  • The term "density" refers to an extensive quality, which means that it is independent of the substance's concentration. Every substance in the world demonstrates its distinctive density. Since it does not fluctuate, it would not affect the rate of melting. The addition of the objects could speed up the process, though, as each one generates heat that could act as the mediating force for the melting process.

To learn more about density, visit :

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1 year ago
Plz i need help fast im timed!!!!!!
professor190 [17]

Answer:

sorry i throght i had the answer

Explanation:

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2 years ago
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Calculate the pressure exerted on the ground by a boy of a mass 60 kg if he stands on one foot.the area of the sole of his shoe
ddd [48]

Answer:

40 Kpa

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150 cm2 = 0.015 m2

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2 years ago
After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg
GuDViN [60]

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses m_1 and m_2 whose respective velocities before collision are u_1 and u_2;

m_1u_1+m_2u_2=m_1v_1+m_2v_2..............(1)

where v_1 and v_2 are their respective velocities after collision.

Given;

m_1=1.3kg\\u_1=5m/s\\m_2=1.3kg\\u_2=0m/s

Note that u_2=0 because the second mass m_2 was at rest before the collision.

Also, since the two masses are equal, we can say that m_1=m_2=m so that equation (1) is reduced as follows;

mu_1+mu_2=mv_1+mv_2\\\\m(u_1+u_2)=m(v_1+v_2)..............(2)

m cancels out of both sides of equation (2), and we obtain the following;

u_1+u_2=v_1+v_2.............(3)

a) When v_1=0.8m/s, we obtain the following by equation(3)

5+0=0.8+v_2\\hence\\v_2=5-0.8\\v_2=4.2m/s

b) As m_1 stops moving v_1=0, therefore,

5+0=0+v_2\\v_2=5m/s

5 0
3 years ago
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