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3 years ago

Water that flows from behind a large dam can cause machines to produce electricity. What change takes place?

1 answer:

8 0

The mechanical energy in the falling water is used to spin the generator, and gets transformed into electrical energy. That's the first choice on the list.

You might be interested in

Multiply.<br> (2x + 4)(x - 4)

sergij07 [2.7K]

Answer:

(2x + 4)(x - 4)=2x^2-4x-16

7 0

3 years ago

In which situation is no work considered to be done by a force?

andre [41]

If the angle is either 0 or 180, that means that there is either negative or positive work, so A and D are not correct.
If the angle is 45, then there is still some work involved.
The only option where there is no work done by a force is B. when the angle is between the force and displacement is 90.

8 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

A geologist sees a fault along which blocks of rock in the footwall have moved higher relative to blocks of rock in the hanging

Luda [366]

Answer: Normal fault

Explanation:

The type of fault that is explained above is a normal fault. We should note that normal faults typically takes place in a divergent boundary in a scenario where the crusts may have been pulled apart.

Since the crust is pulled apart in this case, it leads to the downward movement of the hanging wall which leads to the football being above the hanging wall.

6 0

3 years ago

A pitcher can throw a fastball that reaches home plate at 95 mph. What is this speed in m/s

Taya2010 [7]

42.47 meters per second

5 0

3 years ago