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IRISSAK [1]
3 years ago
11

an electron is released from rest in a region of space with a nonzero electric field.1. As the electron moves, does the electric

potential energy of the system increase, decrease, or stay the same?a. increaseb. decreasec. stay the same2. Choose the best explanation from among the following:a. As the electron begins to move, its kinetic energy increases. The increase in kinetic energy is equal to the decrease in the electric potential energy of the system.b. Because the electron has a negative charge its electric potential energy doesn't decrease, as one might expect, but increases instead.c. The electron will move perpendicular to the electric field, and hence its electric potential energy will remain the same.
Physics
1 answer:
zvonat [6]3 years ago
5 0

Answer:

1. a. increase

2. Because the electron has a negative charge its electric potential energy does not decrease as one might expect, but increases instead.

Explanation:

Lets first consider the relation between the electric field and electric potential.

E = -ΔV/Δs

As this equation indicates that the electric field is due to the change in potential and change in the the position of charge. Electric field is directed towards the decreasing potential and the electron moves in the opposite direction of the electric field  where potential increases. Thats why the best explanation is that the electron has a negative charge it moves towards the positive region where the electric potential energy increases.

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Three 1.5V cells are connected in series in a circuit. What is the total potential difference?
lorasvet [3.4K]

Answer:

4.5V

Explanation:

1.5x3= 4.5

4 0
2 years ago
What does the potential energy diagram of a chemical reaction tell you
sattari [20]
By looking at the potential energies before and after the reaction, we can tell that the reaction is exothermic (final < initial) or endodermic (final > initial).

Also, the amount of activation energy gives an idea of the external energy required to initiate the reaction (for example, by heating the reactants).
Furthermore, by the same principle, we can also deduce the activation energy for the reverse reaction.

If a catalyst is available, the diagram will show a reduced activation energy, compared to a reaction without catalyst.  However, it will also show that the catalyst does not alter the initial and final energies of the reaction.
4 0
3 years ago
Read 2 more answers
A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflect
AleksAgata [21]

Answer:

 A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

           A_res = 2A

           A_resultant = 2 .01

           A_resulting = 0.2 m

8 0
2 years ago
An object moving with a speed of 35 m/a and has a kinetic energy of 1500j, what is the mass of the object
EleoNora [17]

Explanation:

Speed or velocity (V) = 35 m/s

Kinetic energy (K. E) = 1500 Joule

mass (m) = ?

We know

K.E = 1/2 * m * v²

1500 = 1/2 * m * 35²

1500 * 2 = 1225m

m = 3000 / 1225

m = 2.45 kg

The mass of the object is 2.45 kg

Hope it will help :)

5 0
3 years ago
The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of
natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

\mu = Coefficient of friction = 0.4

Energy stored in spring is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J

As the energy in the system is conserved we have

\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s

The speed of the 8 kg block just before collision is 3.258 m/s

7 0
3 years ago
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