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4 years ago

The unit for energy is the

Physics

1 answer:

nikitadnepr [17]4 years ago

4 0

Explanation:

Joule (J) is the MKS unit of energy, equal to the force of one Newton acting through one meter.

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How do power plants generate electricity?

GalinKa [24]

Answer:

Explanation:

electrons are a waste product of bacteria living around plant roots – plants discharge organic matter into the soil, which is broken down by bacteria. It is possible to harvest them using motonless electrodes and turn them into electricity, without affecting the plant’s growth in any way.

4 0

3 years ago

The equation used to predict the theoretical period Ty of a simple pendulum assumes a small amplitude of oscillation. A student

astra-53 [7]

Answer:

The answer is "Choice E".

Explanation:

In this situation the option e is right because its resistance decreases through time, however, the time is the same for the same reason, whereas the sphere deteriorates, somehow it travels shorter distances however if the air resistance becomes are using the amplitude of movement declines, that's why other choices were wrong.

4 0

3 years ago

North Dakota Electric Company estimates its demand trend line (in millions of kilowatt hours) to be: D = 75.0 + 0.45Q, whe

Alborosie

Answer:

The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

Explanation:

Given that,

The demand trend line is

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

We need to calculate the demand forecast for winter

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times101)\times0.80$

$D=96.36\ millions\ KWH$

We need to calculate the demand forecast for spring

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times102)\times1.20$

$D=145.08\ millions\ KWH$

We need to calculate the demand forecast for summer

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times103)\times1.40$

$D=169.89\ millions KWH$

We need to calculate the demand forecast for fall

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times104)\times0.60$

$D=73.08\ millions KWH$

Hence, The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

3 0

3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

4 years ago

In the problem below, what is the student showing?

Phantasy [73]

Answer:

S<em>tudent showing the volume of cube</em>

V = a³

if a = 3 cm

Then volume is a³ = 3 ×3 × 3

= 27 cm³

6 0

3 years ago

The unit for energy is the​

The unit for energy is the