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Marizza181 [45]

3 years ago

11

Simone created a chart to summarize the energy transformations that take place when energy from the wind is used to generate ele

ctricity.
Which best completes the chart?

nuclear energy transformed to electrical energy

chemical energy transformed to electrical energy

radiant energy transformed to mechanical energy

kinetic energy transformed to mechanical energy

2 answers:

Lady bird [3.3K]3 years ago

4 0

Kinetic energy transformed to mechanical energy. The object that is being rotated by the wind gains kinetic energy and the motor, that is connected to the rotating object, gains mechanical energy.

The mechanical energy gained will be transformed into electrical energy once the set up starts to be used for generating electricity.

Naddika [18.5K]3 years ago

3 0

Answer:

d.

Explanation:

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List 3 quantities of waves.

VARVARA [1.3K]

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8 0

3 years ago

Read 2 more answers

To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i

ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

$m_1v_1+m_2v_2=(m_1+m_2)v$

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

$m_1(-v_1)+m_2v_2=(m1+m2)(-v2)$

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

$-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s$

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

$v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s$

hence, the time is t=0.42 s

4 0

3 years ago

An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi

Arturiano [62]

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

$V_x'=0.8c\\V=0.6c\\m=4*10^5kg$

Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

$KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}$

So, to $V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016$ find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
We have an expression from Lorents transformation for relativistic law of addition of velocities as,

$V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V$

Substituting values, we get,

$V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c$

Thus, the KE will be,

$KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J$

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

3 0

2 years ago

A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv

sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

$K_{rel}=(\gamma-1)mc^2$

$\dfrac{K_{rel}}{mc^2}=\gamma-1$

Put the value into the formula

$\gamma=\dfrac{105.7}{0.511}+1$

$\gamma=208$

We need to calculate the electron’s velocity

Using formula of velocity

$\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}$

$\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}$

$\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1$

$v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2$

Put the value into the formula

$v^2=\dfrac{1-(208)^2}{-208^2}\times c^2$

$v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}$

$v=0.9999 c\ m/s$

Hence, The electron’s velocity is 0.9999 c m/s.

6 0

3 years ago

What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on

Wewaii [24]

The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is $252.52\ m/s^2$ .

The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.

The given data:

time, t = 1.1 s

initial speed, u = 1000 km/h = $\frac{2500}{9}\ m/s$

final speed, v = 0 m/s

So we will be using the equation of motion, that is,

v = u + at

$\therefore 0=\frac{2500}{9} + a(1.1)$

$\Rightarrow a=-\frac{2500}{9(1.1)}$

$\therefore a = - 252.52 \ m/s^2$

Hence , the deceleration of the rocket is $252.52\ m/s^2$ .

To learn more about Attention here:

brainly.com/question/28500124

#SPJ4

6 0

1 year ago