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Ivan
2 years ago
8

Which part of a wind-powered system ultimately produces the electricity? A. nacelle B. blade C. turbine D. generator

Physics
2 answers:
djverab [1.8K]2 years ago
5 0
<span>The correct answer is: (D) Generator

Explanation:
In wind-powered systems, the wind energy turns the blades around the rotor of a wind turbine. That rotor is connected to a generator that generates electricity. In other words, the kinectic energy of the wind is converted into electrical energy by using the generator in the wind-powered systems.</span>
abruzzese [7]2 years ago
4 0

Answer:

C) TURBINE

Explanation:

As we know that wind powered system is used to generate electricity using wind energy.

Here wind powered system has different components

1) Blades: these are long plates connected to the turbine. when these plates encountered high speed wind they start rotating on its axle.

This will produce kinetic energy in the axle which is transferred to the turbine.

2) Turbine: It is used to convert mechanical energy of the wind into useful electrical energy.

This is continuously driven by the kinetic energy of wind

So here correct answer is

C) Turbine

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A 1.2 x 103 kg racecar, with a velocity of 8 m/s, collides with an unsuspecting 80 kg honey badger who is standing
aalyn [17]

Answer: 90 m/s

Explanation:

Given

mass of racecar M=1.2\times10^3\ kg

velocity of racecar u=8\ m/s

mass of still honeybadger m=80\ kg

after collision race car is traveling at a speed of v_1=2\ m/s

conserving linear momentum

Mu+m\times0=Mv_1+ mv_2\quad[v_2=\text{velocity of honeybadger after colllision}]

1.2\times10^3\times8+0=1.2\times10^3\times2+80\times v_2

1.2\times10^3(8-2)=80v_2\\v_2=\frac{7.2\times10^3}{80}=90\ m/s

4 0
3 years ago
The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a
bekas [8.4K]

Answer:

The work done on the canister by the 5.0 N force during this time is

54.06 Joules.

Explanation:

Let the initial kinetic energy of the canister be

KE₁ = \frac{1}{2} mv_1^{2} = \frac{1}{2} *3*3.6^{2} = 19.44 J in the x direction

Let the the final kinetic energy of the canister be

KE₂ = \frac{1}{2} mv_2^{2} = \frac{1}{2} *3*7.0^{2} = 73.5 J in the y direction

Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final

= 73.5 J - 19.44 J = 54.06 J

3 0
3 years ago
Read 2 more answers
Which instrument is used to measure liquid volume?
Alecsey [184]

Volumetric cylinders and volumetric flasks

3 0
3 years ago
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Sound waves with a constant frequency of 250 hertz are traveling through air at stp. what is the wavelength of the sound waves
Deffense [45]

Answer:

wave length is 1.2m

Explanation:

since formula of wave length is v/f

v(speed of sound in air at stp is 300ms^-1)

f(frequency 250hertz)

then wave length is 300÷250 which give 1.2m

5 0
1 year ago
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
Read 2 more answers
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