When we add fractions we first check to make sure the ____are the same. Then we ____

The illustration shows a rollercoaster and indicates four different positions the car might be at as it moves along the track. A

Vanyuwa [196]

Answer:

Point a

Explanation:

The potential energy of an object is given by :

P = mgh

m is mass, g is acceleration due to gravity, h is height above ground level.

Potential energy is directly proportional to the position of an object.

In the attached figure, the maximum height is shown at point (a). It means it will have maximum potential energy at a as compared to b,c and d.

5 0

2 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

3 years ago

In which of the fire plume structure zones do the gases accelerate upward?

I am Lyosha [343]

The zone that gases always accelerate upward is the Luminous flame zone. The fire plume is the column of hot gases, flames and smoke rising above a fire. Gases accelerate upward toward the always luminous flame zone. The luminous flame height is the distance between the base of a flame and the point at which the plume is luminous half the time and transparent half the time.

3 0

3 years ago

If 100 kg person weighed 400 N on the planet Zorg,

Brut [27]

Answer:

4 m/s²

Explanation:

Formula

Weight = Mass x Gravity due to acceleration

Solving :

400 N = 100 kg x g
g = 400/100
g = 4 m/s²

5 0

2 years ago

Read 2 more answers

People are constantly being barraged with low-level radiation. Which radiation source below contributes the most low-level radia

Sati [7]

I think the answer is C) Radond

3 0

3 years ago

Read 2 more answers

When we add fractions we first check to make sure the ____are the same. Then we _____only the numerators

When we add fractions we first check to make sure the are the same. Then we _only the numerators