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3 years ago

What energy is a wind turbine transferring>?????????????????????????????????????

Physics

2 answers:

densk [106]3 years ago

7 0

Wind Energy &/or kinetic

Transfer into Electrical

Stella [2.4K]3 years ago

6 0

The wind-mill is taking kinetic energy out of the moving air that spins it, and uses that energy to do things. It may pump water up out of the ground, it may turn a big stone to grind wheat, or it may turn an electric generator to turn the energy it collects into electrical energy.

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A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle

expeople1 [14]

Explanation:

Given that,

Wavelength of the light, $\lambda=691\ nm=691\times 10^{-9}\ m$

(a) Slit width, $a=3.8\times 10^{-4}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}$

$\theta=0.104^{\circ}$

(b) Slit width, $a=3.8\times 10^{-6}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}$

$\theta=10.47^{\circ}$

Hence, this is the required solution.

7 0

4 years ago

The table represents the speed of a car in a northern direction over several seconds. A 2-column table with 5 rows. The first co

Oksanka [162]

Answer:

Column 1 should be titled “Time,” and Column 2 should be titled

“Velocity.” ⇒ 1st answer

Explanation:

* Lets revise some definitions of graphs

A velocity-time graph represents the speed and direction an object

travels over a period of time.

Velocity-time graph is called speed-time graph.

The vertical axis of a velocity-time graph represents the velocity of the

object and horizontal axis represents the time from the initial position

In an acceleration time graph y-axis represents the acceleration, and

x-axis represents the time

The table represents the speed of a car in a northern direction over

several seconds

That means the data in the table represent the velocity of the car in a

certain time

Time (s) Velocity (m/s)

0 5

2 10

4 15

6 20

8 25

10 30

Column 1 would be on the x-axis

x-axis represents the time

Column 2 would be on the y-axis

y-axis represents the velocity

Column 1 should be titled “Time,” and Column 2 should be titled

“Velocity.”

3 0

4 years ago

Read 2 more answers

It took David 2 seconds to lift a 5 Newton bag of toys from the floor to the top of the top shelf, which is 3 meters tall. How m

just olya [345]

So power is equal to work over time and work is force times distance, you do 5 times 3 and get 15 dividing by 2 gives us 7.5 W answer c

7 0

3 years ago

What can you say about the magnitudes of the forces that the balloons exert on each other?

maxonik [38]

Answer:

$F_G=G. \frac{m_1.m_2}{R^2}$ gravitational force

$F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}$ electrostatic force

Explanation:

The forces that balloons may exert on each other can be gravitational pull due to the mass of the balloon membrane and the mass of the gas contained in each. This force is inversely proportional to the square of the radial distance between their center of masses.

The Mutual force of gravitational pull that they exert on each other can be given as:

$F_G=G. \frac{m_1.m_2}{R^2}$

where:

$G=$ gravitational constant $=6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}$

$m_1\ \&\ m_2$ are the masses of individual balloons

$R=$ the radial distance between the center of masses of the balloons.

But when there are charges on the balloons, the electrostatic force comes into act which is governed by Coulomb's law.

Given as:

$F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}$

where:

$\rm \epsilon_0= permittivity\ of\ free\ space$

$q_1\ \&\ q_2$ are the charges on the individual balloons

R = radial distance between the charges.

3 0

3 years ago

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0

4 years ago