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Margarita [4]
3 years ago
15

Suppose you have a device that extracts energy from ocean breakers in direct proportion to their intensity. If the device produc

es 20.0 kW of power on a day when the breakers are 1.20 m high, how much (in kW) will it produce when they are 0.450 m high?
Physics
1 answer:
babymother [125]3 years ago
7 0

Answer:

2.81 kW      

Explanation:

Initial power output, P = 20.0 kW

Initial amplitude, A = 1.20 m

Final Amplitude, A = 0.450 m

Power is proportional to extracted energy which is proportional to the intensity of the wave and intensity of the wave is directly proportional to the square of the amplitude.

P ∝ E ∝ I ∝ A²

\frac{P'}{P}=\frac{A'^2}{A^2}

Substitute the values to find the value of power produced:

P'=\frac{0.450^2}{1.20^2}\times 20.0 kW=2.81 kW

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Compare the magnitude of the electromagnetic and gravitational force between two electrons separated by a distance of 2.00 m. As
Lesechka [4]
First you need to know about two laws, which are:
1) Coulomb's law
2) Newton's law of gravitation

1.
According to Coulomb's law, Electric force between TWO charges is:
F_{e} =  \frac{k*q_{1}*q_{2}}{r^{2}} -- (A)

Where, k = 1/(4*π*epsilon_not) = 9 * 10^{9} \frac{Nm^{2}}{C^{2}}

Both q_{1} and q_{2} = -1.61 x 10^{-19} C
r = Distance between the two charges = 2.00m

Plug-in the above values in (A), you would get:

F_{e} = (9 * 10^{9} ) (1.61 * 10^{-19}  * 1.61 * 10^{-19}) / (2*2)  

F_{e} = 5.83* 10^{-29} N

2.
According to Newton's law of gravitation:
F_{g} =  \frac{Gm_{1}m_{2}}{r^{2}} -- (B)

Where G = Gravitational constant = 6.674 * 10^{-11} m^{2} kg^{-1}s^{-2}

m1 = m2 = Mass of the electron = 9.11 * 10^{-31} kg
r = 2.0 m

Plug-in the above values in (B), you would get:

F_{g} = (6.674 * 10^{-11}) ( 9.11 * 10^{-31}  *  9.11 * 10^{-31}}[/tex]) / (2*2)  

F_{e} = 5.83* 10^{-29} N

F_{e} = 1.38 * 10^{-71} N

Now do Fe over Fg, you would get:
\frac{F_{e}}{F{g}}  = 4.23 * 10^{42}

Ans: So the blanks are:
1) 5.82
2) 1.38
3) 4.23

-i
6 1
3 years ago
Read 2 more answers
Something is thrown in the air and 5.35 seconds later it lands on something 2.1 meters from the ground. How fast was it thrown u
UNO [17]

Answer:

26.6 m/s

Explanation:

Given:

Δy = 2.1 m

t = 5.35 s

a = -9.8 m/s²

Find: v₀

Δy = v₀ t + ½ at²

(2.1 m) = v₀ (5.35 s) + ½ (-9.8 m/s²) (5.35 s)²

v₀ = 26.6 m/s

5 0
3 years ago
A 1.150 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.900 N to
shutvik [7]

k = 5.29

a = 0.78m/s²

KE = 0.0765J

<u>Explanation:</u>

Given-

Mass of air tracker, m = 1.15kg

Force, F = 0.9N

distance, x = 0.17m

(a) Effective spring constant, k = ?

Force = kx

0.9 = k X0.17

k = 5.29

(b) Maximum acceleration, m = ?

We know,

Force = ma

0.9N = 1.15 X a

a = 0.78 m/s²

c) kinetic energy, KE of the glider at x = 0.00 m.

The work done as the glider was moved = Average force * distance

This work is converted into kinetic energy when the block is released. The maximum kinetic energy occurs when the glider has moved 0.17m back to position x = 0  

As the glider is moved 0.17m, the average force = ½ * (0 + 0.9)

Work = Kinetic energy

KE = 0.450 * 0.17

KE = 0.0765J

4 0
3 years ago
Kara is pushing a shopping cart to the right. The applied force is acting _______.
mafiozo [28]

Answer:

D.) to the right

8 0
3 years ago
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Which of these would represent the force of gravity
tigry1 [53]

Missing figure: find it in attachment.

Answer:

Force D

Explanation:

In order to answer the question, let's keep in mind that the force of gravity on an object on Earth is the attractive force exerted by the Earth on the object; its direction is always downward (towards the Earth's centre), and its magnitude is given by

F = mg

where m is the mass of the object and g is the acceleration of gravity.

It follows immediately that in the figure, the force of gravity is the only force acting downward: therefore, force D.

The other forces are called:

Force A: thrust (it is the forward force generated by the engines)

Force B: lift (it is the upward produced by the aerodynamics of the wings)

Force C: air resistance (it is the backward force due to the friction between the air and the surface of the plane)

7 0
3 years ago
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