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Montano1993 [528]
2 years ago
9

running back in a football game ran 5 yards north, then ran 6 yards east before being tackled. What distance did he run?

Physics
2 answers:
vazorg [7]2 years ago
5 0

The player ran 11 yards.

5+6=11

So 11 yards!

Hope dis helps!

-Learning with Natalia~

Maslowich2 years ago
3 0

Answer:

11 yards

Explanation:

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A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl
Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the 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=157^{\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-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

Therfore,

\lambda^{'}-\lambda=4.64 p.m.

Here, the photon's incident wavelength is \lamda=7.33pm

So,

\lambda^{'}=7.33+4.64=11.97 p.m

From the conservation of momentum,

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

here, \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}}

Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

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

8 0
3 years ago
Solid barium sulfate is placed into a beaker to form a saturated solution of barium sulfate. the solution has a barium concentra
borishaifa [10]
Ksp = [Ba⁺²][SO₄⁻²]

[Ba⁺²] = [SO₄⁻²] for barium sulfate
Thus,
Ksp = (1 x 10⁻⁵)²
Ksp = 1 x 10⁻¹⁰
5 0
3 years ago
Read 2 more answers
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IRINA_888 [86]
Sorry don’t know this one
3 0
3 years ago
What is the unique geological feature found on Mercury surface?
Cloud [144]

Answer:

The surface of Mercury has landforms that indicate its crust may have contracted. They are long, sinuous cliffs called lobate scarps. These scarps appear to be the surface expression of thrust faults, where the crust is broken along an inclined plane and pushed upward.

Explanation:

I hope this helps a little bit.

3 0
3 years ago
A 25 kg child stands 2.5 m from the center of a frictionless merry‐go‐round, which has a 200 kg*m^2 moment of inertia and is spi
vodka [1.7K]

Answer:

Explanation:

a ) Time period  T = 2 s

Angular velocity ω = 2π / T

=  2π / 2 = 3.14 rad /s

Initial moment of inertia I₁ = 200 + mr²

= 200 + 25 x 2.5²

=356.25

Final moment of inertia

I₂ = 200 + 25 X 1.5 X 1.5

= 256.25

b ) We apply law of conservation of momentum

I₁ X ω₁ =  I₂ X ω₂

ω₂ = I₁ X ω₁ / I₂

Putting the values

w_2=\frac{356.25\times3.14}{256.25}

ω₂ = 4.365 rad s⁻¹

c ) Increase in rotational kinetic energy

=1/2 I₂ X ω₂² -  1/2 I₁ X ω₁²

.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²

= 684.95 J

This energy comes from work done against the centripetal pseudo -force.

7 0
3 years ago
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