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ELEN [110]
3 years ago
10

The three ropes in the figure are tied to a small, very light ring. Two of these ropes are anchored to walls at right angles wit

h tensions T1 = 20 N and T2 = 60 N. What are the magnitude and direction of the tension vector T3 in the third rope?
Correct: Your answer is correct. N. ° (below the positive x-axis)
Physics
1 answer:
netineya [11]3 years ago
8 0
<span>Let t be the angle made by T3 with the x-axis.

 T3 * cos(t) = 20 N ---- (1)
 T3 * sin(t) = 60 N ----- (2)

 Square both equations and add:

T3^2 = 20^2 + 60^2 = 400 + 3600 = 4000
T3 = sqrt(4000) = 63.25 N
 
divide (2) by (1):
tan(t) = 60/20 = 3
t = arctan(3) = 71.57 degrees.

</span><span>Since the angle clockwise from the x-axis some books may call it a negative angle and say the angle is -71.57 degrees or round it to -71.6 or even -72 degrees.</span>
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3 years ago
Problem 9: Suppose you wanted to charge an initially uncharged 85 pF capacitor through a 75 MΩ resistor to 90.0% of its final vo
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Answer:

t=14.678\times 10^{-3}s

Explanation:

Given:

Capacitance, C = 85 pF = 85 × 10⁻¹² F

Resistance, R = 75 MΩ = 75×10⁶Ω

Charge in capacitor at any time 't' is given as:

Q=Q_o(1-e^{-\frac{t}{RC}})

where,

Q₀ = Maximum charge = CE

E = Initial voltage

t = time

also, Q = CV

V= Final voltage = 90% of E = 0.9E

thus, we have

C\times 0.9E=CE(1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})

or

0.9=1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}

or

e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}=1-0.9=0.1

taking log both sides, we get

ln(e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})=ln(0.1)=ln(\frac{1}{10})

or

-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}=-ln(10)

or

t=75\times 10^6\ \times\ 85\times 10^{-12}\times ln{10}

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The average intensity of light emerging from a polarizing sheet is 0.708 W/m2, and that of the horizontally polarized light inci
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Answer:

Angle θ = 30.82°

Explanation:

From Malus’s law, since the intensity of a wave is proportional to its amplitude squared, the intensity I of the transmitted wave is related to the incident wave by; I = I_o cos²θ

where;

I_o is the intensity of the polarized wave before passing through the filter.

In this question,

I is 0.708 W/m²

While I_o is 0.960 W/m²

Thus, plugging in these values into the equation, we have;

0.708 W/m² = 0.960 W/m² •cos²θ

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Cos θ = 0.8588

θ = Cos^(-1)0.8588

θ = 30.82°

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