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lubasha
1 day ago
6

There is a tree standing in the desert; the sun is rising to the east.

Physics
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A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of
serg [3582]

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We utilize the formula

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the angular momentum magnitude of the person 2 meters prior to jumping onto the merry-go-round?

The equation we apply is

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the angular momentum of the person just before she hops onto the merry-go-round?

We utilize the formula

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular velocity of the merry-go-round after the individual jumps on?

We can apply the Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒  I₀*ω₀ = (I₀ + m*R²)*ω₁

At this point, we can determine ω₁

⇒  ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒  ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒  ω₁ = 0.769 rad/s

5) Once the merry-go-round moves at this new angular speed, what force must the person exert to hold on?

We must calculate the centripetal force as follows

Fc = m*ω²*R  

⇒  Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) When the person is halfway around, they choose to simply release their hold on the merry-go-round to exit the ride.

What is the linear speed of the person the moment they exit the merry-go-round?

we can apply the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular velocity of the merry-go-round after the individual releases their hold?

ω₀ = 1.53 rad/s

It returns to its original angular speed

8 0
1 month ago
A power cycle operates between a lake’s surface water at a temperature of 300 K and water at a depth whose temperature is 285 K.
inna [3103]

Response: a) 0.04 kW = 40 W

b) 0.05

Explanation:

A)

The thermal efficiency of the power cycle is calculated as Input / Output

Input = 10 kW + 14,400 kJ/min which translates to 10 kW + 14,400 kJ/(60s) = 10 kW + 14,400/60 kW.

Output equals 10 kW

Thus, Thermal Efficiency = Output / Input = 10 kW / 250 kW = 0.04 kW = 40 W

B)

Maximum Thermal Efficiency of the power cycle is defined as 1 - T1/T2

where T1 = 285 Kelvin

and T2 = 300 Kelvin

Thus, Maximum Thermal Efficiency = 1 - T1/T2 = 1 - 285/300 = 0.05

8 0
1 month ago
Read 2 more answers
A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. 1.84 x 103 J
Yuliya22 [3333]
<span>The work done corresponds to the potential energy that the person acquires while ascending the stairs.
work = potential energy acquired = mgh
W = 75kg * 9.8m/s² * 2.50m = 1837.5 J</span>
4 0
2 months ago
How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. th
Softa [3030]

Answer:

Refer to the explanation

Explanation:

Solution:-

- For this problem, we will assume the grating has a defined line density ( N ) representing the lines per mm.

- The angle formed by each fringe on the screen is given by ( θ ).

- The order of bright/dark spots is characterized by an integer ( n )

- The incident light's wavelength is ( λ )

- Using the diffraction grating relationship from the Young's Experiment is as follows:

                               n*lambda = \frac{sin(theta)}{N}

- The provided formula corresponds to constructive interference.

- We will examine how increasing the distance between the screen and the grating affects the pattern. Let ( L ) be the distance from the grating center to the screen center. The distance ( yn ) indicates the vertical spacing between each fringe on the screen.

- For small angles ( θ ), we use the approximation of sin ( θ ) ≈ tan ( θ ). Thus,

                            sin ( θ ) ≈ tan ( θ ) = [ yn / L ]

- Replacing this approximation in the diffraction grating relation results in:

                            y_n = n*lamda*L*N

- To double the distance from the screen to the grating, we use the relation with ( 2L ):

                            yn ∝ L

Result: The separation between each order of bright and dark fringe doubles. The interference pattern will spread further! Consequently, there will be fewer bright spots visible on the screen since a larger surface area is required to accommodate the entire pattern. The increased distance also diminishes the intensity contrast between bright and dark fringes due to the extended path traveled by light rays. Intensity is inversely related to the square of the travel distance.

- If the line density of the grating ( N ) were doubled, it follows that:

                            yn ∝ N

Result: The spacing between bright and dark fringes would also double. The interference pattern expands further, leading to the necessity of a bigger screen to show the entire pattern.

4 0
1 month ago
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