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kumpel [21]
2 years ago
8

(1 point) Unknown resistor in voltage divider Suppose that a power supply is connected across two resistors R1 and R2 that are c

onnected in series. The power supply voltage E, the voltage across the second resistor V2, and the first resistance R1 are all known, but R2 is not known. Find an expression for R2 in terms of E,R1,andV2
Physics
1 answer:
just olya [345]2 years ago
5 0

Answer:

R₂ = V₂R₁/(E + V₂)

Explanation:

By voltage divider, V₂ = ER₂/(R₁ + R₂)

So, cross-multiplying, we have

V₂(R₁ + R₂) = ER₂

expanding the bracket, we have

V₂R₁ + V₂R₂ = ER₂

collecting like terms, we have

V₂R₁ = ER₂ + V₂R₂

factorizing, we have

V₂R₁ = (E + V₂)R₂

dividing through by (E + V₂), we have

R₂ = V₂R₁/(E + V₂)

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To solve this problem we will apply the concepts related to wavelength as the rate of change of the speed of the wave over the frequency. Mathematically this is

\lambda = \frac{v}{f}

Here,

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f = Frequency,

Replacing with our values we have that,

\lambda = \frac{340}{500}

\lambda = 0.68m

The distance to move one speaker is half this

\lambda/2 = 0.34m

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3 years ago
A parallel-plate capacitor has square plates that have edge length equal to 1.20×102cm and are separated by 1.00 mm. It is conne
lora16 [44]

Answer:

Energy stored in the capacitor will be 1.835\times 10^{-6}J

Explanation:

We have given edge length of capacitor = 1.2\times 10^2cm=1.2m

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Separation is given as d  1 mm = 10^{-3}m

We know that capacitance is given by C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 1.44}{10^{-3}}=12.744\times 10^{-9}F

Voltage is given as V = 12 volt

We know that energy stored is given by E=\frac{1}{2}CV^2=\frac{1}{2}\times 12.744\times 10^{-9}\times 12^2=1.835\times 10^{-6}J

3 0
3 years ago
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In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
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Answer:

570 N

Explanation:

Draw a free body diagram on the rider.  There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.

The rider is moving at constant speed, so acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F cos 30° - T cos 15° = 0

F = T cos 15° / cos 30°

Sum of the forces in the y direction:

∑F = ma

F sin 30° - W - T sin 15° = 0

W = F sin 30° - T sin 15°

Substituting:

W = (T cos 15° / cos 30°) sin 30° - T sin 15°

W = T cos 15° tan 30° - T sin 15°

W = T (cos 15° tan 30° - sin 15°)

Given T = 1900 N:

W = 1900 (cos 15° tan 30° - sin 15°)

W = 570 N

The rider weighs 570 N (which is about the same as 130 lb).

6 0
3 years ago
Which object is in static equilibrium
balu736 [363]

Answer:

The body is said to be in static equilibrium if the net force acting on a body at rest is zero.As the net force is zero,the body will not undergo motion.

Explanation:

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Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-
torisob [31]

Answer:

6.05×10⁻³ J

Explanation:

Note: Two capacitors connected in series behaves like two resistors connected in parallel.

Using

1/Ct = 1/C1+1/C2

Ct = (C1×C2)/(C1+C2)............................ Equation 1

Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.

Given: C1 = 25 µF, C2 = 50 µF

Substitute into equation 1

Ct = (25×50)/(25+50)

Ct = 1250/75

Ct = 16.67 µF.

Using

Q = CV.................... Equation 2

Where Q = Charge, V = Voltage.

Given: V = 33 V, C = 16.67 µF = 16.67×10⁻⁶ F

Substitute into equation 2

Q = 33(16.67×10⁻⁶)

Q = 5.5×10⁻⁴ C.

Since both capacitors are connected in series, the same amount of charge flows through them.

Using,

E = 1/2Q²/C.................. Equation 3

Where E = Energy stored in the 25-µF capacitor

Given: Q =5.5×10⁻⁴ C, C = 25 µF = 25×10⁻⁶ F

Substitute into equation 3

E = 1/2(5.5×10⁻⁴)²/ 25×10⁻⁶

E = 6.05×10⁻³ J.

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