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3 years ago

How can a wind turbine be used to make an electric circuit?

Physics

1 answer:

Radda [10]3 years ago

7 0

Answer:

Wind turbines operate on a simple principle. The energy in the wind turns two or three propeller-like blades around a rotor. The rotor is connected to the main shaft, which spins a generator to create electricity.

Explanation:

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What is the net charge of a metal ball if there are 21,749 extra electrons in it?

pickupchik [31]

Answer:

$Q=3.47\times 10^{-15}\ C$

Explanation:

Given that,

Number of extra electrons, n = 21749

We need to find the net charge on the metal ball. Let Q is the net charge.

We know that the charge on an electron is $q=1.6\times 10^{-19}\ C$

To find the net charge if there are n number of extra electrons is :

Q = n × q

$Q=21749\times 1.6\times 10^{-19}\ C$

$Q=3.47\times 10^{-15}\ C$

So, the net charge on the metal ball is $3.47\times 10^{-15}\ C$ . Hence, this is the required solution.

6 0

3 years ago

What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?

MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^2}$

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

$I = \frac{100}{4\pi (2.5)^2}$

$I = 1.2738W/m^2$

The relation between intensity I and $E_{max}$

$I = \frac{E_max^2}{2\mu_0 c}$

Here,

$\mu_0$ = Permeability constant

c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

$E_{max}^2 = 2I\mu_0 c$

$E_{max}=\sqrt{2I\mu_0 c }$

$E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)$

$E_{max} = 30.982 V/m$

Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

$B_{max} = \frac{E_{max}}{c}$

$B_{max} = \frac{30.982 V /m}{3*10^8}$

$B_{max} = 1.03275 *10{-7} T$

Therefore the maximum value of the magnetic field is $B_{max} = 1.03275 *10{-7} T$

3 0

3 years ago

A mass attached to a 50.0 cm long string starts from rest and is rotated 40 times in one minute before reaching a final angular

ch4aika [34]

To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

$\theta = \frac{1}{2}\alpha t^2 +\omega_0 t + \theta_0$

Where,

$\theta =$ Angular Displacement

$\alpha =$ Angular Acceleration

$\omega_0 =$ Angular velocity

$\theta_0 =$ Initial angular displacement

For this case we have neither angular velocity nor initial angular displacement, then

$\theta = \frac{1}{2}\alpha t^2$

Re-arrange for $\alpha,$

$\alpha = \frac{2\theta}{t^2}$

Replacing our values,

$\alpha = \frac{2(40rev*\frac{2\pi rad}{1rev})}{60^2}$

$\alpha = 0.139rad/s$

Therefore the ANgular acceleration of the mass is $0.139rad/s^2$

4 0

2 years ago

Captain John Stapp is often referred to as the "fastest man on Earth." In the late 1940s and early 1950s, Stapp ran the U.S. Air

valentina_108 [34]

Answer:

The sled needed a distance of 92.22 m and a time of 1.40 s to stop.

Explanation:

The relationship between velocities and time is described by this equation: $v_f=v_0+a*t$ , where $v_f$ is the final velocity, $v_0$ is the initial velocity, $a$ the acceleration, and $t$ is the time during such acceleration is applied.

Solving the equation for the time, and applying to the case: $t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s} }{-201\frac{m}{s^2} }=1.40s$ , where $v_f=0\frac{m}{s}$ because the sled is totally stopped, $v_0=282\frac{m}{s}$ is the velocity of the sled before braking and, $a=-201\frac{m}{s^2}$ is negative because the deceleration applied by the brakes.

In the other hand, the equation that describes the distance in term of velocities and acceleration: $x_f-x_0=v_0*t+\frac{1}{2}*a*t^2$ , where $x_f-x_0$ is the distance traveled, $v_0$ is the initial velocity, $t$ the time of the process and, $a$ is the acceleration of the process.

Then for this case the relationship becomes: $x_f-x_0=282\frac{m}{s} *1.40s+\frac{1}{2}(-201\frac{m}{s})*(1.40s)^2=94.22m$ .

<u>Note that the acceleration is negative because is a braking process.</u>

4 0

3 years ago

Read 2 more answers

Urgent!

Masja [62]

mass of iron block given as

$m_1 = 1.90 kg$

density of iron block is

$\rho = 7860 kg/m^3$

now the volume of the iron piece is given as

$V = \frac{m}{\rho}$

$V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3$

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

$F_b = \rho_L V g$

here we know that

$\rho_L$ = density of liquid = 916 kg/m^3

$F_b = 916* 2.42 * 10^{-4} * 9.8$

$F_b = 2.17 N$

Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium

$F_s + F_b = mg$

$F_s + 2.17 = 1.90* 9.8$

$F_s = 16.45 N$

So reading of spring balance will be 16.45 N

Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block

$F_n = F_g + F_b$

$F_n = (1 + 2.50)*9.8 + 2.17$

$F_n = 34.3 + 2.17 = 36.47 N$

So the other scale will read 36.47 N

3 0

3 years ago