Which part of a stream's sediment load moves the slowest?

1 answer:

Rudiy273 years ago

4 0

The bed load moves the slowest from all the parts of the stream's sediment. It consists of particles suspended that are suspended and float around the bed. This part is the slowest in motion, as it rolls, and moves with the flow. The particles near the bed are not dissolved so they settle at the bottom and move with the stream.

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An object of mass 6.00 kg falls with an aceleration of 8.00 m/s2. The magnitude of air resitance must be ____ N

allsm [11]

Using Newton's Second Law, we can find the air resistance. We know the net force is equal to mass times acceleration. $F_{net} = m*a = (6.00kg)(8.00 \frac{m}{s^2}) = 48N 

F_{g} - F_{d} = 48N

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3 years ago

A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 385 kV. The secondary of

enot [183]

Answer:

a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new output is 86% of the old output

c) The losses in the new line are 74% the losses in the old line.

Explanation:

a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}$

In the old transformer the ratio of voltages was:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{385} =0.03117\\\\n_2=n_1/0.03117=32.1n_1$

In the new transformer the ratio of voltages is:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{500} =0.024\\\\n_2=n_1/0.24=41.7n_1$

In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new current ratio is

$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{500}= 0.024\\\\I_2=0.024I_1$