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3 years ago

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Consider resistors R1 and R2 connected in series E R1 R2 and in parallel E R1 R2 to a source of emf E that has no internal resis

tance. How does the power dissipated by the resistors in these two cases compare?

1 answer:

mamaluj [8]3 years ago

4 0

Answer:

The power dissipated by the resistors connected in parallel is higher than that of in series.

Explanation:

Power dissipated by a resistor is given by the following formula:

$P = I^2R$

The current, I, can be found by Ohm's Law:

$V = IR$

<u>Case 1: </u>

The equivalent resistance in resistors connected in series is:

$R_{eq} = R_1 + R_2$

$I = V/R = E/(R_1 + R_2)$

$P_1 = I^2R_{eq} = \frac{E^2}{(R_1 + R_2)^2}(R_1 + R_2) = \frac{E^2}{(R_1 + R_2)}$

<u>Case 2:</u>

The equivalent resistance in resistors connected in parallel is:

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\\R_{eq} = \frac{R_1R_2}{R_1 + R_2}$

$I = V/R = \frac{E}{(\frac{R_1R_2}{R_1 + R_2})} = \frac{E(R_1 + R_2)}{R_1R_2}$

$P_2 = I^2R_{eq} = \frac{E^2(R_1 + R_2)^2}{R_1^2R_2^2}\frac{R_1R_2}{(R_1 + R_2)} = \frac{E^2(R_1 + R_2)}{R_1R_2}$

<u>Comparison: </u>

If we compare $P_1$ and $P_2$ :

$P_1 = \frac{E^2}{(R_1 + R_2)}, P_2 = \frac{E^2(R_1 + R_2)}{R_1R_2}\\$

According to these equations, we can conclude that

$P_2 > P_1$

You can give numbers to resistances to compare them.

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Electric resistance ____________ with a(n) ____________in the length of a wire and as a result current flow ___________.

maria [59]

The answer is D. Electric resistance increases with an increase in the length of a wire and as a result current flow decreases. There is a direct relationship between the length of the wire and the resistance. The longer the wire, the more resistance there will be. Additionally, from Ohm's Law, current is inversely proportional to resistance. This means as the current increases, resistance decreases or vice versa.

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3 years ago

A brick of gold is 0.1 m wide, 0.1 m high, and 0.2 m long. The density of gold is 19,300 kg/m3. What pressure does the brick exe

PtichkaEL [24]

Answer:

The pressure exerted by the brick on the table is 18,933.3 N/m².

Explanation:

Given;

height of the brick, h = 0.1 m

density of the brick, ρ = 19,300 kg/m³

acceleration due to gravity, g = 9.81 m/s²

The pressure exerted by the brick on the table is calculated as;

P = ρgh

P = (19,300)(9.81)(0.1)

P = 18,933.3 N/m²

Therefore, the pressure exerted by the brick on the table is 18,933.3 N/m².

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3 years ago

WINSTONCH [101]

Answer:

Its initial position was 471 m.

Explanation:

We have,

Final position of the object is 327 m

Displacement of the object is -144 m

It is required to find its initial position. The difference of final and initial position is equal to the displacement of the object. So,

$d=\text{final position}-\text{initial position}\\\\-144=327-\text{initial position}\\\\\text{initial position}=327+144\\\\\text{initial position}=471\ m$

So, its initial position was 471 m.

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3 years ago

The graph above shows the position and time calculate the velocity of the particle from T=0s to T=4s?

zzz [600]

Answer:

The velocity of the particle from T = 0 s to T = 4 s is;

0.5 m/s

Explanation:

The given parameters from the graph are;

The initial displacement (covered) at time, t₁ = 0 s is x₁ = 1 m

The displacement covered at time, t₂ = 4 s is x₂ = 3 m

The graph of distance to time, from time t = 0 to time t = 4 is a straight line graph, with the velocity given by the rate of change of the displacement to the time which is dx/dt which is also the slope of the graph given as follows;

$The \ slope \ of \ the \ displacement \ time \ graph, \ m =velocity, \ v= \dfrac{x_2 - x_1}{t_2 - t_1}$

$Therefore, \ the \ velocity \ of \ the \ particle \ v = \dfrac{3 \ m - 1 \ m}{4 \ s - 0 \ s} = \dfrac{2 \ m}{4 \ s} = \dfrac{1}{2} \ m/s$

The velocity of the particle from t = 0 s to t = 4 s = 1/2 m/s = 0.5 m/s.

3 0

3 years ago

0.180-kg stone rets on a frictionless, horizontal surface. A bullet of mass 7.50 g, traveling horizontally at 390 m/s, strikes t

Musya8 [376]

Answer:

The magnitud of the velocity is

$8.46m/s$

and the direccion:

$-28.3$ degrees from the horizontal.

Explanation:

Fist we define our variables:

$m_{s}=0.18kg\\m_{b}=7.5g = 0.0075kg\\v_{1b}= 390m/s -i\\v_{1s}=0m/s\\v_{2b}=210m/s -j\\v_{2s}=?$

The letters i and j represent the direction of the movement, i in this case is the horizontal direction, and j is perpendicular to i.

velocities with sub-index 1 are the speeds before the crash, and with sub-index 2 are the velocities after the crash.

Using conservation of momentum:

$m_{s}v_{1s}+m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}\\v_{1s}=0, so\\m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}$

Clearing for the velocity of the stone after the crash:

$v_{2s}=\frac{m_{b}v_{1b}-m_{b}v_{2b}}{m_{s}}$

Substituting known values:

$v_{2s}=\frac{0.0075kg(((390m/s-i)-210m/s-j)}{0.18kg}\\v_{2s}=(16.25m/s-i) - (8.75m/s-j)$

The magnitud of the velocity is :

$|v_{2s}|=\sqrt{(16.25m/s-i)^2 + (8.75m/s-j)^2}\\|v_{2s}|=18.46m/s$

and the direction:

$tan^{-1}(-8.75/16.25)=-28.3$

this is -28.3 degrees from the +i direction or the horizontal direcction.

Note: i and j can also be seen as x and y axis.

3 0

4 years ago