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Damm [24]
3 years ago
7

when a ball is dropped it is easy to see that earth exerts a force on it. why cant you tell that the ball exerts a force on it.

why can't you tell the ball exerts a force on Earth
Physics
1 answer:
inn [45]3 years ago
3 0
Because the distance that each object moves is in inverse proportion to their masses. That means the earth moves such a little tiny tiny tiny bit that you can't see it. but you're right. It's important to understand that when you drop a ball the ball and the earth both move toward each other.
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The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
es la atmósfera que se percibe en un cuento de acuerdo a la manera de pensar y sentir de los personajes no se ve sólo se siente​
masya89 [10]

Answer:

can you please ask in english

8 0
3 years ago
3. When encountering low visibility from rain or fog, you
zimovet [89]

Answer:

c. low beams and fog lights

Explanation:

When encountering low visibility from rain or fog, use your low beams and  fog lights. High beams will only increase the glare. If you can't see at least five seconds in front of you, don't drive. Pull over and put your hazards on until it clears up.

i just got the answer wrong and the drivers ed gave me this explanation !!

8 0
3 years ago
Read 2 more answers
A 500g object falls off a cliff and losers 100 J from its gravitational potential energy store. if the gravitational field stren
ludmilkaskok [199]

Answer : Height, h = 20.4 m

Explanation :

It is given that,

Mass of an object, m = 500 g = 0.5 kg

Gravitational potential energy, PE = 100 J

The Gravitational potential energy is the energy which is possessed due to the height and gravity of an object. It is given as :

PE = m g h

where,

h is the height of the cliff.

100\ J=0.5\ kg\times 9.8\ m/s^2\times h

h = 20.40 m

So, the height of the cliff is 20.4 m.

7 0
3 years ago
A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s
Natasha2012 [34]

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

6 0
3 years ago
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