Answer:
correct option is A. Resonance
Explanation:
solution
vibrations in the tube are caused by the Resonance because Resonance cause of sounds production in the musical instrument
As if resonance ( hollow cylindrical tube) immerse in cylinder of water and force in to vibration by tuning fork
As tines of tuning fork vibrate at natural frequency and it create sound wave that impinged up on opening of resonance tube
so correct option is A. Resonance
Answer:
- 5436 J
Explanation:
mass of car, m = 120 kg
radius of loop, r = 12 m
velocity at the bottom (A) = Va = 25 m/s
Velocity at the top(B) = Vb = 8 m/s
Vertical distance from A to B = diameter of loop, h = 2 x 12 = 24 m
by use of Work energy theorem
Work done by all the forces = change in kinetic energy of the body
Work done by the force + Work done by the friction = Kinetic energy at B - kinetic energy at A
- m x g x h + Work done by friction = 0.5 x 120 x (Vb^2 - Va^2)
- 120 x 9.8 x 24 + Work done by friction = 60 x (64 - 625)
- 28224 + Work done by friction = - 33660
Work done by friction = -33660 + 28224 = - 5436 J
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vo = 25 m/sec
<span>vf = 0 m/sec </span>
<span>Fμ = 7100 N (Force due to friction) </span>
<span>Fg = 14700 N </span>
<span>With the force due to gravity, you can find the mass of the car: </span>
<span>F = ma </span>
<span>14700 N = m (9.8 m/sec²) </span>
<span>m = 1500 kg </span>
<span>Now, we can use the equation again to find the deacceleration due to friction: </span>
<span>F = ma </span>
<span>7100 N = (1500 kg) a </span>
<span>a = 4.73333333333 m/sec² </span>
<span>And now, we can use a velocity formula to find the distance traveled: </span>
<span>vf² = vo² + 2a∆d </span>
<span>0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d </span>
<span>0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d </span>
<span>-625 m²/sec² = (-9.466666666667 m/sec²) ∆d </span>
<span>∆d = 66.0211267605634 m </span>
<span>∆d = 66.02 m</span>
Answer:
A. 
B. 
C. 
D.
Explanation:
Given:
- no. of moles of oxygen in the cylinder,

- initial pressure in the cylinder,

- initial temperature of the gas in the cylinder,

<em>According to the question the final volume becomes twice of the initial volume.</em>
<u>Using ideal gas law:</u>



A.
<u>Work done by the gas during the initial isobaric expansion:</u>




C.
<u>we have the specific heat capacity of oxygen at constant pressure as:</u>

Now we apply Charles Law:



<u>Now change in internal energy:</u>



B.
<u>Now heat added to the system:</u>



D.
Since during final cooling the process is isochoric (i.e. the volume does not changes). So,