All categories

2 years ago

A) the wind is blowing directly towards the wind turbine at a speed of 12 m/s. at this wind

Physics

1 answer:

gulaghasi [49]2 years ago

4 0

The kinetic energy of the air travelling at 12m/s, which passes through

the circular area in 1 second is 5.4 x 10⁵ J.

<h3>What is kinetic energy?</h3>

The kinetic energy of an object is associated with its motion. It can be related to the mass and velocity as

K.E = 1/2 mv²

Given the wind is blowing directly towards the wind turbine at a speed of 12 m/s. At this wind speed, 7500 kg of air passes every second through the circular area swept out by the blades.

Substitute the values, we have

K.E = 1/2 x 7500 x (12)²

K.E = 5.4 x 10⁵ J

Hence, the kinetic energy of the air travelling at 12m/s, which passes through the circular area is 5.4 x 10⁵ J.

Learn more about kinetic energy.

brainly.com/question/12669551

#SPJ1

You might be interested in

A voltmeter was used to check the coolant and a reading of 0.2 volt with the engine off was measured. A reading of 0.8 volt was

Julli [10]

Answer:

C. Technician B

Explanation:

Excessive Galvanic activity:

To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.

Electrolysis problem:

When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.

In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.

7 0

3 years ago

Electromagnetic induction means that moving a magnet through a loop of wire creates an electric current.

MAVERICK [17]

False. An Electromagnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field.

4 0

4 years ago

Read 2 more answers

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

The weight of air measured in units of force per area is called _____.

Vlad1618 [11]

The weight of air resting on a surface, divided by the area
of the surface ... described in units of force per unit area ...
is called air pressure.

6 0

4 years ago

Read 2 more answers

How far can a person run in 15 minutes if he runs at an average speed of 16 km/hr?

anygoal [31]

Answer:

4km

Explanation:

15 minutes is 1/4 of an hour.

1/4 of 16 is 4.

3 0

3 years ago