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solmaris [256]
3 years ago
13

If 28.0 J of work is done by an external force to move a charge from a potential of 8.0 V 8.0 V to a potential of 4.0 V , 4.0 V,

what is the change in electric potential energy?
Physics
1 answer:
jonny [76]3 years ago
5 0

Answer:

Change in electric potential energy is -28.0 J

Explanation:

Electric potential energy is defined as the work is done to move a charge particle from one position to another in space in the presence of other charge particle or electric potential.

                                               OR

Electric potential energy is also equal to the change in the configuration of the charge particles.

Thus,

Change in electric potential energy = - Work Done

According to the problem, Work Done is equal to 28 J. Thus,

Change in electric potential energy = -28 J  

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The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

4 0
2 years ago
A very long straight wire has charge per unit length 1.44×10-10C/m.
4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m            

Explanation:

We have given charge per unit length, that is liner charge density \lambda =1.44\times 10^{-10}C/m

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

E=\frac{\lambda }{2\pi \epsilon _0r}, here \lambda is linear charge density and r is distance of the point where we have to find the electric field

So 2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

3 0
3 years ago
The ozone layer is found in which layer of the atmosphere
prohojiy [21]

This question is incomplete because the options are missing; here is the complete question

The ozone layer is found in which layer of the atmosphere?

A. Stratosphere

B. Mesosphere

C. Thermosphere

D. Troposphere

The correct answer is A. Stratosphere

Explanation:

The ozone layer as indicated by its name is mainly composed of Ozone (O2), this layer is essential for life because it filters ultraviolet radiation and acts as a greenhouse effect gas by trapping part of the heat from the sun. Additionally, the ozone layer is located in the stratosphere, which is the second layer of the atmosphere and can be found between 20 km to 50 km from Earth's surface. Moreover, the existence of the ozone layer in the stratosphere makes the temperature increase with height due to the radiation of the sun filter by ozone.

4 0
2 years ago
The graph shows a ball rolling from A to G. At which point does the ball have the greatest kinetic energy?
AysviL [449]

please give garph picture

3 0
3 years ago
Read 2 more answers
What is nucleic acid
Alekssandra [29.7K]

like dioxyribonucleic acid? if so then that's DNA


8 0
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