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2 years ago

Describe the applications of thermal expansion?

Physics

1 answer:

BartSMP [9]2 years ago

5 0

Answer:

used in fixing iron rim with the wheel firmly

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Please expert solve this

777dan777 [17]

Explanation:

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5 0

2 years ago

A tuba may be treated like a tube closed at one end. If a tuba has a fundamental frequency of 88.4 Hz, determine the first three

yulyashka [42]

Answer with Explanation:

We are given that

Fundamental frequency,f=88.4 Hz

Speed of sound,v=343 m/s

We have to find the first three overtones.

Tube is closed

The overtone of closed pipe is equal to odd number of fundamental frequency.

Therefore, the overtone of tuba

$f'=nf$

Where n=3,5,7,..

Substitute n=3

$f'=3\times 88.4=265.2Hz$

For second overtone

$f'=5\times 88.4=442Hz$

For third overtone

$f'=7\times 88.4=618.8Hz$

4 0

3 years ago

What type of heat transfer is boiling water??

beks73 [17]

Heat can be transferred from one place to another by three methods:

conduction in solids,

convection of fluids (liquids or gases),

radiation through anything that will allow radiation to pass.

6 0

3 years ago

Read 2 more answers

The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular

nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

6 0

3 years ago

A bowling ball is traveling at 3 m/s. It starts to roll up a ramp. How high above the ground will the ball be when it stops roll

Zepler [3.9K]

Answer:

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Explanation:

FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE

8 0

2 years ago

Read 2 more answers