What is the energy conversion in a wind turbine?

A force does work on an object if a component of the force is

IrinaK [193]

Answer:

A force does work on an object if a component of the force is parallel to the displacement of the object.

Explanation:

Work, a measurement of energy is said to be done when a force applied to an object results in the movement of that object to a certain distance and direction. Force is the act of push or pulls occurs on an object as a result of the interaction between that object with another one and displacement is the distance and direction covered by that object as a result of the force applied on it.

The work done (W) by a constant force (F) is equal to the product of the force in the direction of displacement of the object and the distance (d) moved by the object i.e., W = F * d.

The angle between the displacement and the force is θ, then the work done, W = Fd cos θ ........ (1)

Positive work - Force acts in the same direction with respect to the displacement of the object. Here, θ is zero, so cos θ i.e., cos 0 is 1. Therefore, from the equation (1), W = Fd (i.e., work done by the force is positive).

Negative work - Force acts in the opposite direction with respect to the displacement of the object. Here, θ is 180°, so cos θ i.e., cos 180° is -1. Therefore, from the equation (1), W = -Fd (i.e., work done by the force is negative).

If a force is applied to an object and it does not move, then the work done is zero i.e., W = F * 0 = 0. Also, if the force and displacement are at right angle to each other, then θ is 90°. Therefore, from the equation (1), W = 0 since cos 90° is zero.

7 0

4 years ago

Can anyone help me? Physics

olga2289 [7]

I cannot read please show closer

5 0

4 years ago

A student uses a spring to launch a marble vertically in the air. The mass of the marble is

Fed [463]

Answer:

the maximum height reached by the marble is 11.84 m.

Explanation:

Given;

mass of the marble, m = 0.02 kg

extension of the string, x = 0.08 m

force applied to the string, F = 58 N

Apply the principle of conservation of energy;

elastic potential energy = gravitational potential energy

¹/₂fx = mgh

$h = \frac{Fx}{2mg} \\\\h = \frac{58 \ \times \ 0.08}{2 \ \times \ 0.02 \ \times \ 9.8} \\\\h = 11.84 \ m$

Therefore, the maximum height reached by the marble is 11.84 m.

5 0

3 years ago

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

xz_007 [3.2K]

Answer:

a) $\Delta{t} = 5.39s$

b) the motorcycle travels 155 m

Explanation:

Let $t_2-t_1 = \Delta{t}$ , then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

$v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}$

where:

$v_{m2}$ is the speed of the motorcycle at time 2

$v_{c}$ is the velocity of the car (constant)

$v_{0}$ is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

$\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]$

$v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s$

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

$x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933$

3 0

3 years ago

Greg bought 12 bags of dog treats that were each 50 grams. How many kilograms is this?

Alona [7]

Answer: 0.6 kilograms

3 0

3 years ago

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