Ocean waves contain energy that is transferred to them by ______ moving across the surface.a. wind
b. sunlight

Answers

Answer 1

Answer:

Answer

a. wind

Explanation

Ocean waves are coursed by wind which transfers its energy into the water. This forms big waves that can travel a long distance. The size of the wave depend on the speed of wind, the area covered and the duration of the wind.

Answer 2

Answer:

Answer:

Wind

Explanation:

Related Questions

Just after a motorcycle rides off the end of a ramp and launches into the air, its engine is turning counterclockwise at 8325 rev/min. The motorcycle rider forgets to throttle back, so the engine's angular speed increases to 12125 rev/min. As a result, the rest of the motorcycle (including the rider) begins to rotate clockwise about the engine at 4.2 rev/min. Calculate the ratio IE/IM of the moment of inertia of the engine to the moment of inertia of the rest of the motorcycle (and the rider). Ignore torques due to gravity and air resistance.
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Why is the efficiency of a machine always less than 100 percent
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What is velocity of a particle?

Why does cold water have less thermal energy than hot water?

Answers

This is because the water molecules have less kinetic energy.

A long solenoid has 1400 turns per meter of length, and it carries a current of 4.9 A. A small circular coil of wire is placed inside the solenoid with the normal to the coil oriented at an angle of 90.0˚ with respect to the axis of the solenoid. The coil consists of 42 turns, has an area of 1.2 × 10-3 m2, and carries a current of 0.45 A. Find the torque exerted on the coil.

Answers

Answer:

The torque on the coil is $1.955* 10^(- 4)\ N-m$

Solution:

No. of turns per meter length, n = 1400 turns\m

Current, I = 4.9 A

Angle, $\theta = 90.0^(\circ)$

No. of turns of coil, N = 42 turns

Area, A = $1.2* 10^(- 3)m^(2)$

Current in the coil, I' = 0.45 A

Now,

To calculate the exerted torque on the coil:

The magnetic field, B produced inside the coil is given by:

$B = n\mu_(o)I$

$B = 1400* 4\pi times 10^(- 7)* 4.9 = 8.62* 10^(- 3)\ T$

Now, the torque exerted is given by:

$\tau = I'NAB$

$\tau = 0.45* 42* 1.2* 10^(- 3)* 8.62* 10^(- 3) = 1.955* 10^(- 4)\ N-m$

Answer:

$T\approx 1.95* 10^(-4) N.m$

Explanation:

Given:

A long solenoid having

no. of turns per meter, n =1400

current, I = 4.9 A

A small coil of wire placed inside the solenoid

angle of orientation with respect to the axis of the solenoid, $\theta=90\degree$ °

no. of turns in the coil, N = 42

area of the coil, $a= 1.2* 10^(-3) m^2$

current in the coil, $i =0.45 A$

We have for torque:

$T=n.i.a.B. sin\theta$ .......................(1)

∵ $B=\mu_(0) .n.I$ ................................(2)

where:

B= magnetic field

$\mu_0=$ The permeability of free space = $4\pi*10^(-7) T.m.A^(-1)$

Substitute B from eq. (2) into eq. (1) we have:

$T=n.i.a.(\mu_0.N.I ).sin\theta$

putting the respective values in above eq.

$T=42* 0.45* 1.2* 10^(-3)* 4\pi*10^(-7) * 1400* 4.9* sin 90^(\circ)$

$T\approx 1.95* 10^(-4) N.m$

A 70.0 kg man jumping from a window lands in an elevated fire rescue net 11.0 m below the window. He momentarily stops when he has stretched the net by 1.50 m. Assuming that mechanical energy is conserved during this process and that the net functions like an ideal spring, find the elastic potential energy of the net when it is stretched by 1.50 m. (10 pts)

Answers

Answer:

70.15 Joule

Explanation:

mass of man, m = 70 kg

intial length, l = 11 m

extension, Δl = 1.5 m

Let K is the spring constant.

In the equilibrium position

mg = K l

70 x 9.8 = K x 11

K = 62.36 N/m

Potential energy stored, U = 0.5 x K x Δl²

U = 0.5 x 62.36 x 1.5 x 1.5

U = 70.15 Joule

What is required to cause acceleration?

Answers

F = ma

Where m = mass of body, a = acceleration.

F = ma = m(v - u)/t

So for an acceleration to occur, a force is required.

Also a change of velocity is required for acceleration to occur.

50 J of work was performed in 20 seconds. how much power was used to do this task?

Answers

   Power = (amount of work done) / (time to do the work).

   = (50 J) / (20 sec)

   = 2.5 joule/sec = 2.5 watts .

The Correct answer to this question for Penn Foster Students is: 2.5 W

A forklift raises a crate weighing 8.35 × 102 newtons to a height of 6.0 meters. What amount of work does the forklift do?

Answers

You are almost there, often you are given a mass. But, here you already have a force. You just need to use:W=FdWhere W is work done, F is force, d is distance. Units of work will be Nm

Answer:

You just need to use:W=Fd Where W is work done, F is force, d is distance. Units of work will be Nm

Explanation:

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Nonuniform positive acceleration