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kondor19780726 [428]
3 years ago
9

Charges are pushed through an electric circuit by?

Physics
2 answers:
Semmy [17]3 years ago
7 0

Answer:

The particles that carry charge through wires in a circuit are mobile electrons. The electric field direction within a circuit is by definition the direction that positive test charges are pushed. Thus, these negatively charged electrons move in the direction opposite the electric field.

Explanation:

madreJ [45]3 years ago
3 0
  • Electric potential energy is the energy that is used to push charges or electrons among electric circuit
  • The charges which flow in between electric circuit are called mobile electrons
  • The rate of flow is directly proportional to the electric potential energy.
  • Higher the energy higher the flow and vice versa
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In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
Daniel [21]

Answer:

v_2=-133.17m/s, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before (p_i) and after (p_f) the explosion. We will take the east direction as positive.

Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

Which means (since we want v_2 and M=m_1+m_2):

v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

5 0
3 years ago
Divers found two substances on the bottom of the ocean. At room temperature, both substances are liquid. Scientists then transfe
e-lub [12.9K]

Answer:

Substance 2 is a still figure and substance 1 isn't

Explanation:

5 0
2 years ago
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1. How much heat energy ( Q ) is required to heat 2.0 kg of copper from 30.0 oC to 80.0 oC?
djverab [1.8K]

Answer:

Heat capacity, Q = 38500 Joules

Explanation:

Given the following data;

Mass = 2 kg

Initial temperature, T1 = 30°C

Final temperature, T2 = 80°C

Specific heat capacity of copper = 385 J/Kg°C

To find the quantity of heat required;

Mathematically, heat capacity is given by the formula;

Q = mcdt

Where;

Q represents the heat capacity or quantity of heat.

M represents the mass of an object.

C represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 80 - 30

dt = 50°C

Substituting the values into the formula, we have;

Q = 2 * 385*50

Heat capacity, Q = 38500 Joules

4 0
3 years ago
Which refers to the area on the thermometer marked with the letter A? scale bulb mercury strip number line
expeople1 [14]

Answer:

number line

Explanation: hope this helps lol

8 0
3 years ago
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The results of a dart game were precise but not accurate. The accepted value of the game was the center of the dartboard. Which
hoa [83]

Answer:

The darts hit very different areas of the board.

Explanation:

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