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alina1380 [7]
3 years ago
10

High and low tides are the regular pattern of rising and sinking ocean-water levels. They are caused when the sun's gravity and

the moon's gravity pull on ocean water. The moon's gravity has a greater effect on Earth's oceans than the sun's gravity. Why would the moon have a greater effect than the sun?
A. The moon is not as hot as the sun.
B. The moon has less mass than Earth.
C. The moon has less mass than the sun.
D. The moon is closer to Earth than the sun.

Physics
1 answer:
MrRa [10]3 years ago
8 0

Answer:

D. The moon is closer to Earth than the sun.

Explanation:

Tides are formed as a consequence of the differentiation of gravity due to the moon across to the Earth sphere.

Since gravity variate with the distance:

   

F = G\frac{m1\cdot m2}{r^{2}}  (1)                            

Where m1 and m2 are the masses of the two objects that are interacting and r is the distance

For example, see the image below, point A is closer to the moon than point b and at the same time the center of mass of the Earth will feel more attracted to the moon than point B. Therefore, that creates a tidal bulge in point A and point B.

The Sun tidal force contributes to the tidal force of the moon over the earth making high tides higher and low tides lower.  

However, even when the sun is more massive than the moon, it is farther away from the Earth than the moon. So, it is clear by equation 1 that the moon's gravity has a greater effect on Earth's oceans than the sun's gravity.         

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Answer with Explanation:

We are given that

Momentum of photon=8.3\times 10^{-28} kg.m/s

a. We have to find the energy of this photon.

Speed of photon=c=3\times 10^8 m/s

We know that

Momentum=p=\frac{h}{\lambda}

Where

h=6.63\times 10^{-34}J-s=Plank's constant

\lambda=Wavelength of photon

\lambda=\frac{h}{p}

\lambda=\frac{6.63\times 10^{-34}}{8.3\times 10^{-28}}

\lambda=7.99\times 10^{-7} m

E=\frac{hc}{\lambda}

E=\frac{6.63\times 10^{-34}\times 3\times 10^8}{7.99\times 10^{-7}}

E=2.49\times 10^{-19} J

Hence, the energy of photon=2.49\times 10^{-19} J

B.Energy of photon in electron volt=\frac{2.49\times 10^{-19}}{1.6\times 10^{-19}}=1.55 eV

Energy of photon=1.55eV

C.Wavelength of photon =\lambda=7.99\times 10^{-7}m

6 0
3 years ago
Bumper car A (281 kg) moving
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What is the reading shown in millimetre​
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Explanation:

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A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal
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Answer:

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W_g = - \Delta E_p = - (mgh_f  - m g h_i)

where m is the mass of the ball, g is the acceleration due to gravity, h_f is the final height and h_i is the initial height.

So, if the radius is 2.00 m, then the difference of height will be 4 meters:

W_g = - mg (h_f - h_i)

W_g = - 3.00 \ kg \ 9.8 \frac{m}{s^2} \ 4 \m

W_g = - 117.6 Joules

As the tension is perpendicular to the velocity of the ball, the force is always perpendicular to the direction of motion. So, the differential of work will be:

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6 0
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Volgvan

Answer: I think the answer is B

Explanation:

6 0
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