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stepladder [879]
2 years ago
8

While spinning down from 500.0 rpm to rest, a solid uniform flywheel does 5.1 kJ of work. If the radius of the disk is 1.2 m, wh

at is its mass?
Physics
1 answer:
Nookie1986 [14]2 years ago
7 0

Answer:

Mass will be equal to 5.173 kg

Explanation:

Energy due to rotation of uniform flywheel is E = 5.1 KJ = 5100 J

Angular speed \omega =500rpm=\frac{2\times 3.14\times 500}{60}=52.33rad/sec

Radius r = 1.2 m

Rotational kinetic energy of flywheel is E=\frac{1}{2}I\omega ^2

So 5100=\frac{1}{2}\times I\times 52.33 ^2

I=3.724kgm^2

Moment of inertia of solid flywheel is I=\frac{1}{2}mr^2

So 3.724=\frac{1}{2}\times m\times 1.2^2

m=5.173kg

So mass will be equal to 5.173 kg

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A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas
SVEN [57.7K]

Answer:

the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s

Explanation:

given information

car's mass, m = 1200 kg

h_{A} = 100 m

v_{A} = v_{A}

h_{B} = 150 m

v_{B} = 0

according to conservative energy

the distance from point A to B, h = 150 m - 100 m = 50 m

the initial speed v_{A}

final speed  v_{B} = 0

thus,

v_{B}² = v_{A}² - 2 g h

0 = v_{A}² - 2 g h

v_{A}² = 2 g h

v_{A} = √2 g h

    = √2 (9.8) (50)

    = 31.3 m/s

8 0
2 years ago
Calculate a pendulum's frequency of oscillation (in Hz) if the pendulum completes one cycle in 0.5 s.
Marina86 [1]
Time taken to complete one oscillation for a pendulum is Time Period, T = 0.5 s 
Frequency of the pendulum oscillation = 1 / Time Period => f = 1 / T = 1 / 0.5  
Frequency f = 2 Hz
3 0
2 years ago
The nucleus of an atom has all of the following characteristics except that it
Sedaia [141]

Answer:

THE ANSWER IS: contains nearly all of the atom's volume.

Explanation:

3 0
2 years ago
Three wires are made of copper having circular cross sections. Wire 1 has a length l and radius r. Wire 2 has a length l and rad
Alex73 [517]

Explanation:

Below is an attachment containing the solution.

4 0
2 years ago
You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond
luda_lava [24]

Answer:

a) L = 0.75m   f₁ = 113.33 Hz , f₃ = 340 Hz, b) L=1.50m   f₁ = 56.67 Hz ,  f₃ = 170 Hz

Explanation:

This resonant system can be simulated by a system with a closed end, the tile wall and an open end where it is being sung

In this configuration we have a node at the closed end and a belly at the open end whereby the wavelength

With 1  node         λ₁ = 4 L

With 2 nodes      λ₂ = 4L / 3

With 3 nodes       λ₃ = 4L / 5

The general term would be      λ_n= 4L / n         n = 1, 3, 5, ((2n + 1)

The speed of sound is

         v = λ f

         f = v / λ

         f = v  n / 4L

Let's consider each length independently

L = 0.75 m

        f₁ = 340 1/4 0.75 = 113.33 n

         f₁ = 113.33 Hz

        f₃ = 113.33   3

       f₃ = 340 Hz

L = 1.5 m

       f₁ = 340 n / 4 1.5 = 56.67 n

       f₁ = 56.67 Hz

       f₃ = 56.67 3

       f₃ = 170 Hz

8 0
2 years ago
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