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Vera_Pavlovna [14]
3 years ago
9

One circuit contains only an ac generator and a resistor, and the rms current in this circuit is IR. Another circuit contains on

ly an ac generator and a capacitor, and the rms current in this circuit is IC. The maximum, or peak, voltage of the generator is the same in both circuits and does not change. If the frequency of each generator is tripled, by what factor does the ratio IR/IC change?
Physics
1 answer:
snow_tiger [21]3 years ago
7 0

Answer:

1/3 times.

Explanation:

Let V₀ be the peak voltage .

IR ( rms )  = ( V₀ / √2 R )

R is value of resistance

IC =  ( V₀ ωC / √2  )

( 1 / ωC is reactance of capacitance in ac circuit  )

\frac{I_R}{I_C} =\frac{\frac{V_0}{\sqrt{2}R } }{\frac{V_0\omega C}{\sqrt{2} } }

= \frac{I_R}{I_C} = \frac{1}{\omega C R}

When frequency is tripled angular frequency will also be tripled

hence the ratio \frac{I_R}{I_C} becomes 1/3 times.

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Specific gravity for liquids is nearly always measured with respect to water at its densest

Explanation:

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8 0
2 years ago
The weight of a box is 1200 Newton if it exerts a pressure of 800 Pascal calculate the area​
fomenos

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  1. 1.5m2

Explanation:

P=F/A. So here the force is given and the pressure is also given so you make the area the subject since that is what u are looking for

5 0
3 years ago
Charge is flowing through a conductor at the rate of 420 C/min. If 742 J. of electrical energy are converted to heat in 30 s., w
Y_Kistochka [10]

Answer:

3.53 V

Explanation:

Electric charge: The is the rate of flow of electric charge along a conductor.

The S.I unit of electric charge is C.

Mathematically it is expressed as,

Q = It ............................ Equation 1

Where Q = electric charge, I = current, t = time.

I = Q/t.......................... Equation 2

From the question, charge flows through the conductor at the rate of 420 C/mim

Which means in 1 min, 420 C of charge flows through the conductor.

Hence,

Q = 420 C, t = 1 min = 60 seconds

Substitute into equation 2

I = 420/60

I =7 A

Also

P = VI......................... Equation 3

Where P = power, V = potential drop, I = current.

V = P/I................... Equation 4

Note: Power = Energy/time

From the question, P = 742/30 = 24.733 W. and I = 7 A.

Substitute these values into equation 4

V = 24.733/7

V = 3.53 V

Hence the potential drop across the conductor =  3.53 V

4 0
3 years ago
Powerful sports cars can go from zero to 25 m/s (about 60 mph) in 6 seconds. What is the magnitude of the acceleration, includin
Vikentia [17]

Answer:

The magnitude of the acceleration is 4.2 m/s²

Explanation:

Given;

initial velocity of the powerful sports car, u = 0

final velocity of the powerful sports car, v = 25 m/s

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Acceleration = Δv / Δt

Δv is change in velocity

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The magnitude of the acceleration of the powerful sports car during the motion is given by;

a = \frac{v-u}{t}\\\\a = \frac{25-0}{6}\\\\a = 4.2 \ m/s^2

Therefore, the magnitude of the acceleration is 4.2 m/s²

7 0
3 years ago
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