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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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AnnZ [28]

Answer:

\boxed{v=\frac {V}{\sqrt {2}}}

Explanation:

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Now, the speed of the father will be V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}}) while for the child the speed will be v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})

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\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}

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1 year ago
A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by
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Answer:

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Explanation:

Length of sample, l_{s} = 2 cm

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Mobility of the minority carriers ( electrons), \mu_{n} = \frac{V_{d} }{E}

Where the drift velocity, V_{d} = \frac{L}{t_{d} }

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and the Electric field strength, E = \frac{V}{l_{s} }

E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec

The electron diffusion coefficient, D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }

\triangle x = (\triangle t )V_{d}, where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3}  }\\\triangle x =0.533 cm

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For the Einstein equation to be satisfied, \frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V

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4 0
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Answer:

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Answer :

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Explanation:

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Converting to g/m^3,

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Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 × 10^23 × (10.35 × 10^6)/107.87

= 5.778 × 10^28 atoms/m^3

The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6

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