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Marina86 [1]
2 years ago
6

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

uces 10.0 kW of power on a day when the breakers are 1.20 m high, how much will it produce when they are 0.600 m high?
Physics
2 answers:
tamaranim1 [39]2 years ago
6 0

Answer:

It will produce 2.5 Kw at 0.6m high

Explanation:

We are given;

Initial Power output of device; P_i = 10 Kw

Initial amplitude; A_i = 1.2m

Final Amplitude; A_f = 0.6m

We know that power is directly proportional to energy because

P = Energy(work done)/time taken

Thus; P ∝ E - - - - (eq1)

Now,from the question, we are told that Energy is proportional to the intensity. Thus;

E ∝ I - - - - (eq2)

where I is intensity

Now, from formula of Intensity, which is; I = (1/2)(ρ²•ⲕω²•A²)

We can see that I is directly proportional to square of Amplitude A²

Thus, I ∝ A² - - - - (eq3)

Combining eq 1,2 and 3,we can deduce that;

P ∝ E ∝ I ∝ A²

Thus, P ∝ A²

Now, let's set up the proportion as;

P_i/P_f = A_i²/A_f²

Since we are looking for final power, let us make P_f the subject.

So,

P_f = (P_i•A_f²)/A_i²

Plugging in the relevant values to obtain ;

P_f = (10 x 0.6²)/1.2² = 2.5 Kw

slava [35]2 years ago
3 0

Answer:

4.988kW

Explanation:

According to the question, energy E extracted from the ocean breaker is directly proportional to the intensity I. It can be expressed mathematically as E ∝ I

E = kI where k is the constant of proportionality.

From the formula; k = E/I

This shows that increase in energy extracted will lead to increase in its intensity and vice versa.

If the device produces 10.0 kW of power on a day when the breakers are 1.20 m high

E = 10kW and I = 1.20m

k = 10/1.20

k = 8.33kW/m

To know how much energy E that will be produced when they are 0.600 m high, we will use the same formula

k = E/I where;

k = 8.33kW/m

I = 0.600m

E = kI

E = 8.33 × 0.6

E = 4.998kW

The device will produce energy of 4.998kW when they are 0.600m high.

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AleksAgata [21]

Answer:

μ = 0.408

Explanation:

given,

speed of the automobile (u)= 20 m/s

distance = 50 m

final velocity  (v) = 0 m/s

kinetic friction = ?

we know that,

v² = u² + 2 a s

0 = 20² + 2 × a × 50

a = \dfrac{400}{2\times 50}

a = 4 m/s²

We know

F = ma = μN

ma = μ mg

a = μ g

\mu = \dfrac{a}{g}

\mu = \dfrac{4}{9.81}

μ = 0.408

hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408

5 0
2 years ago
A 2400-kg satellite is in a circular orbit around a planet. the satellite travels with a constant speed of 6670 m/s. the radius
Ad libitum [116K]
The gravitational force exerted on the satellite is called the centrifugal force, the force keeping it orbiting to the planet. Its formula is F= mass times the square of the velocity all over the radius.Thus,

F = 2400 * 6670^2 * (1/8.92x10^6) 
F = 11,970 N

I hope I was able to help you. Have a good day.
4 0
2 years ago
Read 2 more answers
As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road i
Kitty [74]

Answer:

<em>B</em><em>.</em><em> </em><em>Kinetic</em><em> </em><em>friction</em><em> </em>

Explanation:

This is definitely the correct answer because kinetic friction acts when an object is in motion and it allows the object to move without slipping, etc

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6 0
2 years ago
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The length of a wire 2.00 m is measured as 2.02m. What is the percentage error in the measurement?
n200080 [17]

Answer:

1%

Explanation:

Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.

Percent Error=\frac{V_{measured}- V_{true} } {V_{true}} *100

The measured value is 2.02 meters and the actual value is 2.00 meters.

V_{measured}=2.02\\\\V_{true}=2.00

Percent Error=\frac{2.02-2.00}{2.00} *100

First, evaluate the fraction. Subtract 2.00 from 2.02

Percent Error=\frac{0.02}{2.00}*100

Next, divide 0.02 by 2.00

PercentError=0.01 *100

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6 0
2 years ago
A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0° below the horizontal. It is held in position by a fri
liubo4ka [24]

Answer:

T = 7.64 kN

F_y = 0.52 kN(Downwards)

F_x = 3.23 kN (Towards Left)

Explanation:

As we know that beam is in equilibrium

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Now by torque balance equation at the pivot we can say

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As we know that

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