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rusak2 [61]
2 years ago
6

Three sticks are arranged to form a right triangle. If the lengths of the three sticks are 0.47 m, 0.62 m and 0.78 m, what are t

he three angles of the triangle?
Physics
1 answer:
Svetradugi [14.3K]2 years ago
6 0

The angles in the triangle are 91 degrees, 53 degrees and 36 degrees respectively.

<h3>What is the cosine rule?</h3>

From the cosine rule we know that;

c^2 = a^2 + b^2 - 2abcosC

Since;

a = 0.47 m

b =  0.62 m

c =  0.78 m

Then;

(0.78)^2 = (0.47)^2 + (0.62)^2 - 2(0.47 * 0.62)cosC

0.61 = 0.22 + 0.38 - 0.58 cosC

0.61 - ( 0.22 + 0.38) = - 0.58 cosC

0.01 = - 0.58 cosC

C = cos-1(0.01/-0.58)

C = 91 degrees

Using the sine rule;

b/Sin B = c/Sin C

0.62/sinB = 0.78/sin 91

0.62/Sin B = 0.78

B = sin-1 (0.62//0.78)

B = 53 degrees

Angle A is obtained from the sum of angles in a triangle;

180 - (91 + 53)

A = 36 degrees

Learn more about triangle:brainly.com/question/2773823

#SPJ1

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Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

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Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.

C₅ = \frac{4.25\times3.98 }{4.25 + 3.98  }

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Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3  = 35.46 μC

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Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = \frac{Q}{C_{4} } = \frac{35.46\times10^{-6} }{3.98\times10^{-6}} = 8.90 volts

Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.

Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC

Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC

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