<h2>Potential energy lost by 10 N rock will be greater</h2>
Explanation:
Two rocks of 5N and 10N falls from the same height . Thus they will loose the potential energy.
The potential energy lost = mass x acceleration due to gravity x height
The potential energy lost by first 5 N rock = 5 h
Because weight of rock m g = 5 N
Similarly , the potential energy lost by 10 N Rock = 10 h
here weight of rock m g = 10 N
Thus comparing these two , the potential energy lost by 10 N rock is greater than that of 5 N rock .
Answer:
Explanation:
Total momentum of the system before the collision
.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left
If v be the velocity of the stuck pucks
momentum after the collision = 2 v
Applying conservation of momentum
2 v = - .75
v = - .375 m /s
Let after the collision v be the velocity of .5 kg puck
total momentum after the collision
.5 v + 1.5 x .231 = .5v +.3465
Applying conservation of momentum law
.5 v +.3465 = - .75
v = - 2.193 m/s
2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.
Kinetic energy before the collision
= 2.25 + 1.6875
=3.9375 J
kinetic energy after the collision
= .04 + 1.2 =1.24 J
So kinetic energy is not conserved . Hence collision is not elastic.
3 ) Change in the momentum of .5 kg
1.5 - (-1.0965 )
= 2.5965
Average force applied = change in momentum / time
= 2.5965 / 25 x 10⁻³
= 103.86 N
The fundamental frequency of the tube is 0.240 m long, by taking air temperature to be 
C is 367.42 Hz.
A standing wave is basically a superposition of two waves propagating opposite to each other having equal amplitude. This is the propagation in a tube.
The fundamental frequency in the tube is given by
where, 
Since, T=37+273 K = 310 K
v = 331 m/s
Using this, we get:
Hence, the fundamental frequency is 367.42 Hz.
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Answer:
R (120) = 940Ω
Explanation:
The variation in resistance with temperature is linear in metals
ΔR (T) = R₀ α ΔT
where α is the coefficient of variation of resistance with temperature, in this case α = -0,0005 / ºC
let's calculate
ΔR = 1000 (-0,0005) (120-0)
ΔR = -60
Ω
ΔR = R (120) + R (0) = -60
R (120) = -60 + R (0)
R (120) = -60 + 1000
R (120) = 940Ω
