All categories

2 years ago

What is the potential drop across R3? (R1 = 20 Ω, R2 = 40 Ω, R3= 60 Ω, V = 60 V) (Ohm's law: V = IR)

1 answer:

7 0

Correct answer: Option D. 30 V

The resistors are connect in series, because in parallel the potential drop across all resistors will have been the same and will be equal to the battery voltage i.e. 60 V.

First we need to find the equivalent resistance and the current passing through the circuit.

Since, resistors are in series, equivalent resistance will be the sum of all individual resistances. So,

Equivalent Resistance = 20 + 40 + 60 = 120 ohm
Current through the circuit = Voltage / Equivalent Resistance = 60/120 = 0.5 ampere

The voltage drop across R3 =Resistance of R3 x Current
= 60 x 0.5
= 30 V

Therefore, D option is the correct answer.

You might be interested in

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago

On sunday a strong gale blows near beth’s house. the best estimate for the wind speed is ____ miles/hour.

Rufina [12.5K]

Answer:

B. 47-54 miles/hour

Explanation:

Gale is a strong wind which is depicted by red warning flag. According to U.S. National Weather Service gale is a sustained surface wind. It is also used to refer winds from tropical coastal areas.

On the basis of force of wind gale is divided in four groups:

Near gale - 32-38 mph
Gale - 39-46 mph
Strong gale - 47-54 mph
Storm - 55-63 mph

8 0

3 years ago

A series circuit has two 10-ohm bulb is added in a series. Technician A says that the three bulbs will be dimmer than when only

ANTONII [103]

Answer:

Technician A is right. The situation will happens even with only two bulbs in series

Explanation:

We must take into account that

1.- All electric device need its nominal voltage to operate

2.-Any and all electric device means an electric load for the source in terms of equation that means any device will implies a drop voltage of V = I*R ( I the flows current and R the resistance of the device)

3.-Nominal voltage for bulbs are specify for houses voltages you find between fase and neutral wires for instance in Venezuela 120 (v).

4.-In a imaginary circuit of only one bulb, the nominal voltage will be applied and the bulb will operates correctly, but when you add another bulb (in series) the nominal voltage will split between the two bulbs ( we could find a situation such as the first bulb work properly but the second one does not). The voltage split according to Ohms law (in such way that the sum of voltage between the terminal of the first bulb plus the voltage at terminals of the second one are equal to nominal voltage.

For that reason all the bulbs are connected in parallel in wich case all of them will operate with the common voltage

4 0

3 years ago

A rope is used to pull a 4.36 kg block at constant speed 4.33 m along a horizontal floor. The force on the block from the rope i

geniusboy [140]

Answer:

Explanation:

The horizontal component of force applied = 7.97 cos 26.2 = 7.15 N.

Vertical component in upward direction = 7.97 sin 26.2 = 3.52 N.

Since the body is moving with uniform velocity, friction force will equalize the external horizontal component = 7.15 N.

So frictional force = 7.15 N.

a) Work done by rope force per second = force in horizontal direction x displacement per second = 7.15 x 4.33 = 30.95 J

b) Increase in thermal energy per second will be due to negative work done by frictional force = work done by external force = 30.95 J.

c) Normal force acting downwards = weight - vertical component of external force = 4.36 x 9.8 - 3.52 = 39.21 N

coefficient of friction = friction force / normal force = 7.15 / 39.21 = 0.18

4 0

3 years ago

The Coulomb force between two charges q1 and q2 at separation r in the air is 10N. If half of the separation is filled with medi

adoni [48]

Answer:

The value of new coulomb force is 1.43 N.

Explanation:

Given;

Coulomb's force in vacuum (air), $F_v$ = 10 N

dielectric constant, K = 7

The Coulomb's force between two charges separated by a distance r in a vacuum is given as;

$F_v = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2}$

The Coulomb's force between two charges separated by a distance r in a medium with dielectric constant is given as

$F_m = \frac{1}{4\pi K\epsilon_0} \frac{q_1q_2}{r^2}$

Take the ratio of the two forces;

$\frac{F_v}{F_m} = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2} \ \times \ \frac{4\pi K\epsilon_0 r^2}{q_1q_2} = K\\\\\frac{F_v}{F_m} = K\\\\\frac{10}{F_m} = 7\\\\F_m = \frac{10}{7} \\\\F_m = 1.43 \ N$

Therefore, the value of new coulomb force is 1.43 N.

5 0

2 years ago