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olya-2409 [2.1K]

3 years ago

10

The power dissipated in each of two resistors is the same. The current across resistor A is triple that across resistor B. If th

e resistance of resistor B is R, what is the resistance of A?

1 answer:

Svetllana [295]3 years ago

5 0

Answer:

Explanation:

this is the answer to your question

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What is the equivalent resistance?

mote1985 [20]

Answer:

C 3.33 Ω

Explanation:

From the equation given 1 / Rtot = 1/10 + 1/5 = 3/10

then Rtot = 10/3 = 3.33 Ω

As an aside:

When there are only two resistors in parallel, the equivalent R is

= R1*R2 / (R1 + R2) = 10*5 / (10+5) = 50 / 15 = 3.33 Ω

7 0

2 years ago

At what distance does a 100 Watt lightbulb deliver the same power per unit surface area as a 75 Watt lightbulb produces 10 m awa

ch4aika [34]

Answer:

At 11.5 m

Explanation:

The power per unit area corresponds to the intensity, which is given by

$I=\frac{P}{4\pi r^2}$

where

P is the power

$4\pi r^2$ is the area irradiated at a distance r from the source (it corresponds to the surface area of a sphere of radius r)

Here we want the intensity of the two light bulbs to be the same, so

$I_1 = I_2\\\frac{P_1}{4 \pi r_1^2}=\frac{P_2}{4\pi r_2^2}$

where we have

P1 = 100 W is the power of the first light bulb

P2 = 75 W is the power of the second light bulb

r2 = 10 m is the distance from the second light bulb

Solving for r1, we find

$r_1 = r_2 \sqrt{\frac{P_1}{P_2}}= (10 m) \sqrt{\frac{100 W}{75 W}} = 11.5 m$

4 0

2 years ago

A series combination of two resistors, 7.25 ω and 4.03 ω, is connected to a 9.00 v battery.

almond37 [142]

a. $11.28\Omega$

The equivalent resistance of a series combination of two resistors is equal to the sum of the individual resistances:

$R_{eq}=R_1 + R_2$

In this circuit, we have

$R_1 = 7.25 \Omega\\R_2 = 4.03 \Omega$

Therefore, the equivalent resistance is

$R_{eq}=7.25 \Omega + 4.03 \Omega=11.28 \Omega$

b. 5.8 V, 3.2 V

First of all, we need to determine the current flowing through each resistor, which is given by Ohm's law:

$I=\frac{V}{R_{eq}}$

where V = 9.00 V and $R_{eq}=11.28 \Omega$ . Substituting,

$I=\frac{9.00 V}{11.28 \Omega}=0.8 A$

Now we can calculate the potential difference across each resistor by using Ohm's law again:

$V_1 = I R_1 = (0.8 A)(7.25 \Omega)=5.8 V$

$V_2 = I R_2 = (0.8 A)(4.03 \Omega)=3.2 V$

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What type of wave travels through matter only??

Olenka [21]

A mechanical wave can only travel through matter.

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