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Nikitich [7]
3 years ago
6

(a) You wish to determine the height of the smokestack of a local coal burning power plant. You convince a member of the mainten

ance crew to mount the support for a simple pendulum at the top of the stack and you suspend a 1.00 kg mass that just misses the ground at its lowest point from the pendulum cord. If the period of the pendulum is 18.7 s, determine the height of the smokestack. 8455.69 Incorrect: Your answer is incorrect. What factors influence the period of a simple pendulum
Physics
1 answer:
kicyunya [14]3 years ago
5 0

Answer:

a. 86.80 m

b. i. The mass of the bob

ii. The length of the pendulum

Explanation:

a. Determine the height of the smokestack.

Using T = 2π√(L/g) where T = period of pendulum = 18.7 s, L = length of pendulum = height of smokestack and g = acceleration due to gravity = 9.8 m/s².

So, making L subject of the formula, we have

T = 2π√(L/g)

T/2π = √(L/g)

squaring both sides, we have

(T/2π)² = L/g

L = (T/2π)²g

Substituting the values of the variables into the equation, we have

L = (T/2π)²g

L = (18.7 s/2π)²(9.8 m/s²)

L = (2.976 s)²(9.8 m/s²)

L = 8.857 s² × 9.8 m/s²

L = 86.796 m

L ≅ 86.80 m

b. What factors influence the period of a simple pendulum

The factors that influence the period of a simple pendulum are

i. The mass of the bob

ii. The length of the pendulum

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v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

<u>Explanation:</u>

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2

Substituting the velocities in the equation

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

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Answer:

572.3 nm

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n_{oil} = refractive index of the oil film = 1.48

t_{oil} = thickness of the oil film = 290 nm

\lambda = wavelength of the dominant color

m = order

Using the equation

2 n_{oil} t_{oil} = (m + 0.5) \lambda

For m = 0

2 (1.48) (290) = (0 + 0.5) \lambda

\lambda = 1716.8 nm

For m = 1

2 (1.48) (290) = (1 + 0.5) \lambda

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For m = 2

2 (1.48) (290) = (2 + 0.5) \lambda

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Hence the dominant color wavelength is 572.3 nm

8 0
4 years ago
A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia
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For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

E_{t} = KE_{i} + PE_{i}

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

   

2) The <em>potential energy</em> at point A is:

PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

KE_{A} + PE_{A} = KE_{B} + PE_{B}

KE_{B} = KE_{A} + PE_{A} - PE_{B}

Since

KE_{A} + PE_{A} = KE_{i} + PE_{i}

we have:

KE_{B} = KE_{i} + PE_{i} - PE_{B} =  19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J

5) The <em>kinetic energy</em> of the roller coaster at point C is:

KE_{i} + PE_{i} = KE_{C} + PE_{C}            

KE_{C} = KE_{i} + PE_{i} = 19820 J      

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

KE_{C} = \frac{1}{2}mv_{C}^{2}

v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

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I hope it helps you!

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3 years ago
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Galina-37 [17]

Answer:

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