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joja [24]
3 years ago
11

An AC generator with a variable frequency and an RMS voltage output of 120 Volts is connected in series to a 3.1 μF Capacitor an

d a 100 Ω resistor. The RMS voltage across the capacitor is measured to be 92 Volts. What is the RMS voltage across the resistor?
Physics
1 answer:
Maksim231197 [3]3 years ago
7 0

Answer:

The RMS voltage across the resistor = 28 V

Explanation:

Capacitor: A capacitor is an electrical device that has the ability to store electrical charges in an electrical circuit. It is expressed in Farad (F)

Resistor: A resistor is an electrical device that oppose the flow of electric current in a circuit. It is expressed in ohms (Ω)

RMS Voltage : RMS voltage  value of an alternating voltage is defined as that value of steady voltage which would dissipate heat at the same rate in a given resistance

Since the it is a series circuit, the total voltage is divided across the resistance and the capacitor.

Vt = V₁ + V₂...........................Equation 1

Where Vt = total Rms voltage = 120 V ,  V₁ = Rms voltage across the Capacitor = 92 V, V₂ = Rms voltage across the resistor.

Making V₂ the subject of the equation in equation 1 above,

V₂ = Vt - V₁  = 120 - 92

V₂ = 28 V.

The RMS voltage across the resistor = 28 V

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The final velocity of the tool is 107.4 m/s

Explanation:

We can solve this problem by using the principle of conservation of energy.

In fact, if air resistance is negligible, the total mechanical energy of the tool is conserved during the fall, so we can write:

K_i + U_i = K_f + U_f

where

K_i = 0 is the kinetic energy of the tool at the top (zero since it is at rest)

U_i = mgh is the gravitational potential energy of the tool at the top, where

m is the mass of the tool

g is the acceleration of gravity

h is the heigth of the tool

K_f = \frac{1}{2}mv^2 is the kinetic energy of the tool just before hitting the ground, where

v is the final speed of the tool

U_f = 0 is the gravitational potential energy of the tool at the bottom (zero since the height is zero)

Re-arranging the equation,

mgh=\frac{1}{2}mv^2

where we have

g=9.8 m/s^2\\h=588 m

And solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8)(588)}=107.4 m/s

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3 0
3 years ago
calculate minimum number of incident photons per area and of the minimum dose needed to visualize an object of 1 mm squared usin
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Answer:

minimum number of photon is 4.05 × 10^{7}

Explanation:

given data

energy = 50 keV = 50 × 10^{3} eV =  50 × 10^{3} × 1.602× 10^{-19}  J

thickness = 10^-3

contrast = 1%

to find out

number of incident photons

solution

we know here equation that is

E  = n  × h  × ν   .......................1

put here all these value

50 × 10^{3} = n × 6.6× 10^{-34} × c/ 1× 10^{-3}

50 × 10^{3} × 1.602× 10^{-19}  = n × 6.6× 10^{-34} ×( 3 × 10^{8} / 1× 10^{-3})

solve it and find n

n = 4.05 × 10^{7}

so here minimum number of photon is 4.05 × 10^{7}

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Moving the probe 1 cm towards the non-grounded electrode changes the value the potential from about 0.90 V to about 1.2 V. Expla
svp [43]

Answer:

-30 N/C

Explanation:

Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m

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Since E = -ΔV/Δx

substituting the values of the variables into the equation, we have

E = -ΔV/Δx

E = -0.30 V/0.01 m

E = -30 V/m

Since 1 V/m = 1 N/C.

E = -30 N/C

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