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dusya [7]
3 years ago
10

In this circuit the battery provides 3 V, the resistance R1 is 7 Ω, and R2 is 5 Ω. What is the current through resistor R2? Give

your answer in units of Amps. An Amp is 1 Coulomb of charge flowing through a cross-sectional area of the wire per second - that's a lot of charge per second and will warm up a typical wire quite a bit! Most devices have circuits with larger resistors - kLaTeX: \OmegaΩ (103 LaTeX: \OmegaΩ) and MLaTeX: \OmegaΩ (106 LaTeX: \OmegaΩ) are common.

Physics
1 answer:
sveta [45]3 years ago
3 0

Answer:

The current pass the R_2 is  I  = 0.25 A

Explanation:

The diagram for this question is shown on the first uploaded image  

From the question we are told that

    The voltage  is  V =  3V

     The first resistance is  R_1 = 7 \Omega

     The second resistance is  R_2 = 5 \Omega

Since the resistors are connected in series their equivalent resistance is  

       R_{eq} =  R_1 +R_2

Substituting values

         R_{eq} = 7 + 5

         R_{eq} = 12 \Omega

Since the resistance are connected in serie the current passing through the circuit  is the same current passing through R_2 which is mathematically evaluated as

        I  =  \frac{V}{R_{eq}}

Substituting values  

      I  =  \frac{3}{12}

      I  = 0.25 A

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A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge Q as shown in the following figure. The particle, confined to move along the x axis, is moved a small distance x along the axis ( where x << a) and released. Show that the particle oscillates in simple harmonic motion with a frequency given by,

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2 years ago
A 425 g block is released from rest at height h0 above a vertical spring with spring constant k = 460 N/m and negligible mass. T
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Answer:

(a) = +5.38m (b) = -5.38m (c) = 1.246m (d) = +0.3771m.

Explanation:

Initially the spring is at equilibrium,

Work done by all forces = change in kinetic energy

Work = ∇K.E

Work = Kf -Ki =0

Since the work done = 0 since the body is at rest.

W(spring) + W(gravity) = 0

W(spring) + W(gravity) = 0

W(spring) = -W(gravity)

Work done by the block on the spring = W(block/spring)

W(block/spring) + W(spring) = 0

W(spring) = -∫kx.dx

W(spring) = ½k(X²i - X²f) ; Xi =0, Xf = 15.3cm = 0.153m

W(spring) = -½* 460 * (0.153)²

W(spring) = - 5.38NM

Work done by block on spring = + 5.38NM

(b). Workdone by spring on the block = -5.38NM.

Note: This is so because the displacement of the force is in the opposite direction to the previous one since they counter each other to maintain equilibrium.

(C). W(spring) +W(gravity) = 0

½kx² + mg(h + x) = 0

-5.83 + mg(h + 0.153) =0

5.83 = 0.425*9.8 (h + 0.153)

5.83 = 4.165(h + 0.153)

H = 1.399 - 0.153

H = 1.246m

(D).

If the release height was 6ho

H = 6* 1.246m = 7.476m

W(spring) = W(gravity)

½kx² = mg(7.476 + x)

Note: At maximum compression, the blocks would be at rest.

½Kx² = mg(h + x)

½ * 460 * x² = 0.425 * 9.8 * (7.476 + x)

230x² = 4.165 (7.476 + x)

230x² = 31.137 + 4.165x

230x² - 4.165x - 31.137 = 0

Solving the quadratic equation ( i would suggest you use formula method for easy navigation of the variables)

X = + 0.3771m or -0.3589m

But we can't have a negative compression value,

X = + 0.3771m

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