Answer:
(A) 0.63 J
(B) 0.15 m
Explanation:
length (L) = 0.75 m
mass (m) =0.42 kg
angular speed (ω) = 4 rad/s
To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)
I = Ic + m
Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis
h is the horizontal distance between the center of mass and the rotational axis of the rod
I = ![(\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2}](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B12%7D%29%28mL%5E%7B2%7D%20%29%20%2B%20m%28%5Btex%5D%5Cfrac%7BL%7D%7B2%7D)
)^{2}[/tex]
I = ![(\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2}](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B12%7D%29%280.42%20x%200.75%5E%7B2%7D%20%29%20%2B%20%28%200.42%20x%20%28%5Btex%5D%5Cfrac%7B0.75%7D%7B2%7D)
)^{2}[/tex])
I = 0.07875 kg.m^{2}
(A) rods kinetic energy = 0.5I
= 0.5 x 0.07875 x 
= 0.63 J 0.15 m
(B) from the conservation of energy
initial kinetic energy + initial potential energy = final kinetic energy + final potential energy
Ki + Ui = Kf + Uf
at the maximum height velocity = 0 therefore final kinetic energy = 0
Ki + Ui = Uf
Ki = Uf - Ui
Ki = mg(H-h)
where (H-h) = rise in the center of mass
0.63 = 0.42 x 9.8 x (H-h)
(H-h) = 0.15 m
Answer:
The buoyant force is 3778.8 N in upward.
Explanation:
Given that,
Mass of balloon = 222 Kg
Volume = 328 m³
Density of air = 1.20 kg/m³
Density of helium = 0.179 kg/m³
We need to calculate the buoyant force acting
Using formula of buoyant force
Where, 
= density of air
V = Volume of balloon
g = acceleration due to gravity
Put the value into the formula
This buoyant force is in upward direction.
Hence, The buoyant force is 3778.8 N in upward.
Answer:
D
Explanation:
Gene flow is the transfer of genetic variation from one population to another
The change in potential energy when the block falls to ground is -480J.
The maximum change in kinetic energy of the ball is 480 J.
The initial kinetic energy of the ball is 0 J.
The final kinetic energy of the ball is 0.148J.
The initial potential energy of the ball is 0.187 J.
The final potential energy of the ball is 0 J.
The work done by the air resistance is 0.039 J.
<h3>Change in potential energy when the block falls to ground</h3>
ΔP.E = -mgh
ΔP.E = -Wh
ΔP.E = - 40 x 12
ΔP.E = -480 J
<h3>Maximum change in kinetic energy of the ball</h3>
ΔK.E = - ΔP.E
ΔK.E = - (-480 J)
ΔK.E = 480 J
<h3>Initial kinetic energy of the ball</h3>
K.Ei = 0.5mv²
where;
- v is zero since it is initially at rest
K.Ei = 0.5m(0) = 0
<h3>Final kinetic energy</h3>
K.Ef = 0.5mv²
K.Ef = 0.5(0.0091)(5.7)²
K.Ef = 0.148 J
<h3>Initial potential energy of the ball</h3>
P.Ei = mghi
P.Ei = 0.0091 x 9.8 x 2.1
P.Ei = 0.187 J
<h3>Final potential energy</h3>
P.Ef = mghf
P.Ef = 0.0091 x 9.8 x 0
P.Ef = 0
<h3>Work done by the air resistance</h3>
W = ΔE
W = P.E - K.E
W = 0.187 J - 0.148 J
W = 0.039 J
Learn more about potential energy here: brainly.com/question/1242059
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<h3 />
Answer:
The amplitude of vibration of string will increase due to which loudness of sound will increase
Explanation:
As we know that the guitar is based on the principle of Resonance. When string of the guitar vibrates at a given frequency then the sound produced in the hollow part of the guitar will also be at same frequency.
This is known as resonance condition, so guitar will produce same frequency sound as that of frequency of string.
Now if the string is plucked with increasing level of force then it will increase the amplitude of vibrations of the string due to which the sound produced in the guitar will also be of same level.
So here we can say that amplitude and intensity of sound related as
so on increasing amplitude the intensity will increase and hence it will produce loud sound
