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Simora [160]
3 years ago
12

A mass of 2.82 kg is hung from a spring, causing the spring to stretch 0.331 m. If a second mass of 3.09 kg is now added to the

spring, how far with it now stretch with both masses hanging?? State your answer to the correct number of significant digits and include the proper units.
Physics
1 answer:
irina1246 [14]3 years ago
4 0

Answer:

0.694 m

Explanation:

Case 1 : When only mass of 2.82 kg is hanged from spring

m = mass hanged from the spring = 2.82 kg

x = stretch caused in the spring = 0.331 m

k = spring constant

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

k (0.331) = (2.82) (9.8)

k = 83.5 N/m

Case 2 : When both masses are hanged from spring

m = mass hanged from the spring = 3.09 + 2.82 = 5.91 kg

x = stretch caused in the spring = ?

k = spring constant = 83.5 N/m

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

(83.5) x = (5.91) (9.8)

x = 0.694 m

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Answer:

(a) 47.08°

(b) 47.50°

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Angle of incidence  = 78.9°

<u>For blue light : </u>

Using Snell's law as:

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Θ₁ is the angle of incidence

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n₁ is the refractive index of air which is 1

So,  

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{sin\theta_2}=0.7323

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<u>For red light : </u>

Using Snell's law as:

\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}

Where,  

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₂ is the refractive index for red light which is 1.331

n₁ is the refractive index of air which is 1

So,  

\frac {sin\theta_2}{sin{78.9}^0}=\frac {1}{1.331}

{sin\theta_2}=0.7373

Angle of refraction for red light = sin⁻¹ 0.7373 = 47.50°.

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Question 1
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