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raketka [301]
3 years ago
6

Power windmills turn in response to the force of high-speed drag. For a sphere moving through a fluid, the resistive force, FR i

s proportional to r 2 v 2 , where r is the radius of the sphere and v is the fluid speed. The power developed, P = FR v, is proportional to r 2 v 3 . The power developed by a windmill can be expressed as P = d r2 v 3 , where r is the windmill radius, v is the wind speed, and d = 4.3 W s3 /m5 . For comparison, a typical home needs about 3.0 kW of electric power. Note: This representation ignores system efficiency, which is about 25%. For a home windmill with a radius of 1.59 m, calculate the power delivered to the generator if the wind velocity is 12.4 m/s.\
Physics
1 answer:
Crazy boy [7]3 years ago
3 0

Answer:

The answer is 20727w

Explanation:

The formula is below;

P = d r^2 v^3 *efficiency

In the question, it is stated that the registration ignores efficiency so we are going to ignore efficiency in the equation and use it this way;

P = d r^2 * v^3

d =4.3, r = 1.59, v =n 12.4

Therefore, P = 4.3 X 1.59^2  X 12.4^3 = 20727W

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1 year ago
Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iro
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3 years ago
Marking branliest!
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D)evaluating a solution

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A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If
Salsk061 [2.6K]

Complete question:

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?

Answer:

The current in the circuit 7 ms later is 0.2499 A

Explanation:

Given;

Ideal inductor, L = 45-mH

Resistor, R =  60-Ω

Ideal voltage supply, V = 15-V

Initial current at t = 0 seconds:

I₀ = V/R

I₀  = 15/60 = 0.25 A

Time constant, is given as:

T = L/R

T = (45 x 10⁻³) / (60)

T = 7.5 x 10⁻⁴ s

Change in current with respect to time, is given as;

I(t) = I_o(1-e^{-\frac{t}{T}})

Current in the circuit after 7 ms later:

t = 7 ms = 7 x 10⁻³ s

I(t) = I_o(1-e^{-\frac{t}{T}})\\\\I =0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}}})\\\\I = 0.25(0.9999)\\\\I = 0.2499 \ A

Therefore, the current in the circuit 7 ms later is 0.2499 A

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Answer:

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Explanation:

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