All categories

2 years ago

How does a wind turbine transform mechanical energy into electrical energy?

Physics

1 answer:

Alex777 [14]2 years ago

4 0

Answer:

"A turbine takes the kinetic energy of a moving fluid, air in this case, and converts it to a rotary motion. As wind moves past the blades of a wind turbine, it moves or rotates the blades. These blades turn a generator."

You might be interested in

R: Law of Conservation of Energy Pre-Write

Bad White [126]

Answer:

V at C is 3.6 m/s

Explanation:

At A kinetic energy is zero and potential energy=mgh=0.5*9.81*0.6=2.943 J

By conservation of energy.

KE+PE=Constant

At C PE=0.6 J

the KE=2.943-0.6=2.343 J

KE=0.5*m*v^2

v=√[KE/(0.5*m)]=3.06 m/s

3 0

3 years ago

Find the acceleration of the system and the tension in the ropes for the system shown. The table mass is 30 kg and the hanging m

marusya05 [52]

The system's tension is 616 N and acceleration is 5.6 $m / s^{2}$

<u>Explanation:</u>

From newton’s second law of motion which state that net force acting on a body is product of mass of a body and acceleration of a body which is given as,

$F_{n e t}=m_{t o t} \times a$

Where,

$F_{n e t}$ is net force acting on body

$m_{\mathrm{tot}}$ is mass of body

a is acceleration of body

Given values

Table mass (m) = 30 kg

Hanging mass (m) = 40 kg

$a=\frac{F_{n e t}}{m_{\mathrm{tot}}}=\frac{m \times g}{m_{\mathrm{tot}}}$

Put the value for m = hanging mass = 40 kg and $g=9.8 \mathrm{m} / \mathrm{s}^{2}$ , we get

$a=\frac{40 \times 9.8}{30+40}=\frac{392}{70}=5.6 \mathrm{m} / \mathrm{s}^{2}$

The tension in the ropes, $T=(m \times g)+(m \times a)$

Here, m as hanging mass

T = tension, N or $k g m / s^{2}$

m = mass, kg

g = gravitational force, $9.8 \mathrm{m} / \mathrm{s}^{2}$

a = acceleration, $m / s^{2}$

$T = (40 \times 9.8)+(40 \times 5.6) = 392+224 = 616 N$

3 0

2 years ago

Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

3 years ago

A 600 N astronaut travels to an asteroid where the gravitational force is one-hundredth that of Earth. What is the astronaut's w

Ganezh [65]

Answer:

A) 6N

Explanation:

the weight of the astronaut on earth can be calculated with the formula

Weight = mass * gravity of earth

Since the gravitational force is one-hundredth of Earth, the formula should be

Weight = mass * (gravity of earth / 100)

Weight = (mass * gravity of earth) / 100

Since you already know the weight, you only need to divide by 100

Weight = (mass * gravity of earth) / 100

Weight = 600N / 100

Weight = 6N

4 0

2 years ago

The battleship and enemy ships 1 and 2 lie along a straight line. Neglect air friction. battleship 1 2 Consider the motion of th

DaniilM [7]

Answer:

$\frac{t_1}{t_2} = \frac{sin\theta_1}{sin\theta_2}$

Explanation:

The vertical component of the initial velocities are

$v_v = v_0sin\theta$

If we ignore air resistance, and let g = -9.81 m/s2. The the time it takes for the projectiles to travel, vertically speaking, can be calculated in the following motion equation

$v_vt - gt^2/2 = s = 0$

$t(v_v - gt/2) = 0$

$v_v - gt/2 = 0$

$t = 2v_v/g = 2v_0sin\theta/g$

So the ratio of the times of the flights is

$t_1 / t_2 = \frac{2v_0sin\theta_1/g}{2v_0sin\theta_2/g} = \frac{sin\theta_1}{sin\theta_2}$

8 0

2 years ago