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3 years ago

Q. At what point in a waterfall do the drops of water contain the most kinetic energy ?

Physics

1 answer:

antoniya [11.8K]3 years ago

7 0

Answer:

2.When they reach the bottom of the fall

Explanation:

The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).

Thus, the correct option is "2" When they reach the bottom of the fall

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A solenoidal coil with 26 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 20.0 cm long

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Answer:

Part a)

$\phi = 2.76 \times 10^{-7} T m^2$

Part B)

$M = 5.52 \times 10^{-5} H$

Part C)

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Explanation:

Part a)

Magnetic field due to a long ideal solenoid is given by

$B = \mu_0 n i$

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$n = 1750 turn/m$

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$B = (4\pi \times 10^{-7})(1750)(0.100)$

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Now magnetic flux due to this magnetic field is given by

$\phi = B.A$

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$\phi = (2.2 \times 10^{-4})(\pi(0.02)^2)$

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Part B)

Now for mutual inductance we know that

$\phi_{total} = M i$

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3 years ago

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Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

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ΔP.E = ΔK.E

mb*g*H = 0.5*mb*vb^2

vb = √2*g*H

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M2 = mo*vo*(L/2)

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M1 = M2

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ΔP.E = ΔK.E

mo*g*h = 0.5*mo*vo^2

h = vo^2 / 2*g

h = 16.14604^2 / 2*(9.81)

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