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3 years ago

How long is a football field?

Physics

2 answers:

Nana76 [90]3 years ago

7 0

Answer:

91.44 meters or 300 feet

ycow [4]3 years ago

6 0

Answer:

100 yards

Explanation:

If comparing from actual playing field between the goal posts. Add another 10 yards for end lines. Making it 120 yards in length.

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A 6kg rock rolls down a hill with a momentum of 12kg m/s . Work out the velocity of the rock

Nat2105 [25]

Answer:

if you need to get the work=force×displacement or if you need the velocity=displacement÷time taken

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3 years ago

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Calculation on simple pendulum

kiruha [24]

Answer:State the question so I can answer

Explanation:

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3 years ago

Calculate the internal axial load at a point D if length L=7 ft. The part is subjected to loads P1=632 lbs, P2=888 lbs (applied

liraira [26]

Answer:

- 256 lbs

Explanation:

The internal axial load at point D can be calculated as the change in the subjected loads. if the magnitude of the horizontal direction = zero

$EF_x = 0$ ; Then:

internal axial load at point D = Δ P

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3 years ago

A red rubber ball rolls down a hill from rest with an acceleration of 7.8 m/s 2 . How fast is it moving after it has traveled 5

masya89 [10]

Answer:

39 m/s

Explanation:

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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