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Cerrena [4.2K]
3 years ago
6

Three lamps with 40 Ω, 30 Ω, and 15 Ω resistance are connected in parallel to a 90 volt circuit. What is the total current used

by the three lamps?
A) 0.94 A


B) 15.25 A


C) 11.25 A


D) 1.06 A
Physics
1 answer:
Lena [83]3 years ago
8 0

Answer: Option C) 11.25 A

Explanation:

Given that:

Voltage V = 90 volts

total current I = ?

Three resistors each with 40 Ω, 30 Ω, and 15 Ω in parallel, the equivalent resistance of the combination (Rtotal) of the circuit is as follows:

i.e 1/Rtotal = (1/R1 + 1/R2 + 1/R3)

1/Rtotal = (1/40Ω + 1/30Ω + 1/15Ω)

1/Rtotal = (3 + 4 + 8)/120Ω

1/Rtotal = 15/120Ω

1/Rtotal = 1/8Ω

To get the value of Rtotal, cross multiply

Rtotal x 1 = 8Ω x 1

Rtotal = 8Ω

Then, apply the formula V = I x Rtotal

90 volts = I x 8Ω

I = 90 volts / 8Ω

I = 11.25 Amperes

Thus, the total current used by the three lamps is 11.25 Amperes

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<h2>Answer:</h2>

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(b) 4.752 x 10⁶ J

<h2>Explanation:</h2>

(a) The given charge (Q) is 110 A·h (ampere hour)

Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;

=> Q = 110A·h

=> Q = 110 x 1A x 1h          [1 hour = 3600 seconds]

=> Q = 110 x A x 3600s

=> Q = 396000A·s

=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C

Therefore, the number of coulombs of charge is 3.96 x 10⁵C

(b) The energy (E) involved in the process is given by;

E = Q x V           -----------------(i)

Where;

Q = magnitude of the charge = 3.96 x 10⁵C

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Substitute these values into equation (i) as follows;

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8 0
3 years ago
At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f
jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

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Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

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Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

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The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

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This question can be solved using the concept of friction energy.

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Learn more about friction energy here:

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Centripetal Force = [mass * (velocity)^2]/Radius
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