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3 years ago

How does a longshore current change the beach

1 answer:

5 0

Longshore current brings about the transportation of sediments on the beach. Since waves usually approach the shore are an angle, the beach sand are carried up or down depending on the direction of the waves resulting in beach drift, that is, the net movement of the beach sand.

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A sled with a mass of 25.0 kg rests on a horizontal sheet of essentially frictionless ice. It is attached by a 5.00-m rope to a

Tpy6a [65]

Answer:

The value is $F = 34.3 \ N$

Explanation:

From the question we are told that

The mass of the shed is $m = 25 \ kg$

The length of the rope is $l = 5.0 \ m$

The angular speed of the shed is $w = 5 rev/minute = \frac{2* \pi * 5}{60 }= 0.524 \ rad/sec$

Generally the force exerted on the shed is mathematically represented as

$F = m * w^2 * l$

=> $F = 25 * 0.524^2 * 5$

=> $F = 34.3 \ N$

7 0

3 years ago

A current of 12 amps is measured in a circuit with a total resistance of 9.0 ohms. What is the size of the voltage source that s

Lynna [10]

A. 108 volts is the answer.

7 0

3 years ago

Read 2 more answers

What is required for an electromagnet to produce a magnetic field that is

jek_recluse [69]

Answer:

A. A coil of wire with current running through it

3 0

3 years ago

Read 2 more answers

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electr

Nata [24]

Answer:

- E = 11.55J

- Q = 0.17C

- E' = (1/4)E

Explanation:

- To calculate the amount of energy stored in the capacitor, you use the following formula:

$E=\frac{1}{2}CV^2$

C: capacitance = 3800.0*10^-6F

V: potential difference = 78.0V

$E=\frac{1}{2}(3800.0*10^{-6}C)(78.0V)^2=11.55J$

The energy stored in the capacitor is 11.55J

- If the electrical energy stored in the capacitor is 6.84J, the charge on the capacitor is:

$E=\frac{1}{2}QV\\\\Q=\frac{2E}{V}\\\\Q=\frac{2(6.84J)}{78.0V}=0.17C$

The charge on the capacitor is 0.17C

- If you take the capacitor as a parallel plate capacitor, you have that the energy stored on the capacitor is:

$E=\frac{1}{2}CV^2=\frac{1}{2}(\frac{\epsilon_oA}{d})V^2=\frac{1}{2}\frac{\epsilon_oAV^2}{d}\\\\$

A: area of the plates

d: distance between plates

If the distance between plates is increased by a factor of 4, you have:

$E'=\frac{1}{2}\frac{\epsilon_oAV^2}{(4d)}=\frac{1}{4}\frac{\epsilon_oAV^2}{2d}=\frac{1}{4}E$

Then, the stored energy in the capacitor is decreased by a a factor of (1/4)

6 0

3 years ago

What is the magnetic field amplitude of an electromagnetic wave whose electric field amplitude is 10v/m?

Nitella [24]

Answer:

$3.33\cdot 10^{-8} T$

Explanation:

For an electromagnetic wave, the relationship between magnetic field amplitude and electric field amplitude is given by

$E=cB$

where

E is the amplitude of the electric field

c is the speed of light

B is the amplitude of the magnetic field

For the electromagnetic wave in this problem, we have

E = 10 V/m is the amplitude of the electric field

So if we solve the formula for B, we find the amplitude of the magnetic field:

$B=\frac{E}{c}=\frac{10 V/m}{3\cdot 10^8 m/s}=3.33\cdot 10^{-8} T$

4 0

3 years ago