All categories

3 years ago

A bus contains a 1500 kg flywheel (a disk that has a 0.600 m radius) and has a total mass of 10,000 kg.

Physics

1 answer:

Westkost [7]3 years ago

4 0

Answer:

Explanation:

moment of inertia of flywheel = 1/2 m R²

= .5 x 1500 x .6²

= 270 kg m²

If required angular velocity be ω

rotational kinetic energy = 1/2 I ω²

= .5 x 270 x ω² = 135 ω²

kinetic energy of bus when its velocity is 20 m/s

= 1/2 x 10000 x 20²

= 2000000 J

Given 90 % of rotational kinetic energy is converted into bus's kinetic energy

135 ω² x 0.9 = 2000000 J

ω²=16461

ω = 128.3 radian /s

b )

Let the height required be h .

Total energy of bus at the top of hill = mgh + 1/2 m v²

m ( gh + .5 v²)

= 10000 ( 9.8h + .5 x 3²)

From conservation of mechanical energy theorem

10000 ( 9.8h + .5 x 3²) = 2000000

9.8h + .5 x 3² = 200

9.8h = 195.5

h = 19.95 m .

You might be interested in

Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i

leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are $\dfrac{\lambda}{2}$

Thus; for both speakers; the wavelength of the sound is:

$\dfrac{\lambda}{2} = (10+30) cm$

$\dfrac{\lambda}{2} = (40) cm$

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{8}+ \Delta \phi_o$

$\pi \ rad = \dfrac{ \pi}{4}+ \Delta \phi_o$

$\Delta \phi_o = \pi -\dfrac{ \pi}{4}$

$\Delta \phi_o = \dfrac{ 4\pi - \pi}{4}$

$\Delta \phi_o = \dfrac{ 3\pi}{4} \ rad$

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad$

$\Delta \phi = \dfrac{3 \pi}{4} \ rad$

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

$A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)$

$A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)$

$A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)$

A = 0.765a

7 0

3 years ago

What are the variables in Gay-Lussac's law? pressure and volume pressure, temperature, and volume pressure and temperature volum

Yanka [14]

Answer:

pressure and temperature (assuming volume is constant)

Explanation:

3 0

3 years ago

Read 2 more answers

In the combo circuit diagrammed, R1 = 19.2 Ω, R2 = 20.7 Ω, and R3 = 25.8 Ω. Find the equivalent resistance of the circuit.

garik1379 [7]

Answer:

Equivalent resistance: 13.589 Ω

Explanation:

R series = R1 + R2 + R3 ...

$\frac{1}{R_{eq} } = \frac{1}{R1} +\frac{1}{R2} +\frac{1}{R3} ...$

Find the equivalent resistance of the right branch of the circuit:

$R_{eq} = R_{2} +R_{3} \\R_{eq} = 20.7 + 25.8 = 46.5 ohms$

$\frac{1}{R_{eq} } = \frac{1}{19.2} +\frac{1}{46.5}\\\\\frac{1}{R_{eq} } = 0.0735887097\\\\R_{eq} = 13.5890411$

6 0

2 years ago

A bus travels east for 3 km, then north for 4 km. What is its final displacement?

ikadub [295]

5 km northeast. Left and up would make northeast

8 0

3 years ago

A person is lying on a diving board 2.00 m above the surface of the water in a swimming pool. She looks at a penny that is on th

wlad13 [49]

Answer:

7.98 m

Explanation:

In the given question,

distance above surface= 2 m

Distance penny from person = 8 m

Since the swimming pool is filled with water and atmosphere has air therefore the refractive index phenomenon will occur.

The refractive index of water: air is 4/3 (1.33).

Using the formula, 4/3 = real depth, apparent depth

real depth= 4/3 x apparent depth

Now, calculating apparent depth = 8 - 2

= 6 m

therefore, real depth = 4/3 x apparent depth

= 1.33 x 6

= 7.98

thus, 7.98 m is the real depth of water.

8 0

3 years ago