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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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g Suppose that you charge a 3 F capacitor in a circuit containing eight 3.0 V batteries, so the final potential difference acros
Vladimir79 [104]

The complex, highly technical formula for capacitors is

<em>Q = C V</em>

Charge = (capacitance) (voltage)

Charge = (3 F) (24 V)

<em>Charge = 72 Coulombs</em>

The positive plate of the capacitor is missing 72 coulombs worth of electrons.  They were sucked into positive terminal of the battery stack.

The negative plate of the capacitor has 72 coulombs worth of extra electrons.  They came from the negative terminal of the battery stack.

You should be aware that this is a humongous amount of charge !  An average <u><em>lightning bolt</em></u>, where electrons flow between a cloud and the ground for a short time, is estimated to transfer around <u><em>15 coulombs</em></u> of charge !

The scenario in the question involves a "supercapacitor".  3 F is is no ordinary component ... One distributor I checked lists one of these that's able to stand 24 volts on it, but that product costs $35 apiece, you have to order at least 100 of them at a time, and they take 2 weeks to get.  

Also, IF you can charge this animal to 24 volts, it will hold 864J of energy.  You'd probably have a hard time accomplishing this task with a bag of leftover AA batteries.

7 0
3 years ago
A proton is released in a uniform electric field, and it experiences an electric force of 5 X 10^-10 N toward the South. What is
Ymorist [56]

Answer:

Electric field on proton

E=3.12\times 10^9\ N/C

Explanation:

Given that

Force,F=5\times 10^{-10}\ N

We know that

Charge on proton

q=1.6\times 10^{-19}\ C

We know that

Force = Electric field x Charge

F= E x q

E=\dfrac{F}{q}\ N/C

E=\dfrac{5\times 10^{-10}}{1.6\times 10^{-19}}\ N/C

E=3.12\times 10^9\ N/C

Electric field on proton

E=3.12\times 10^9\ N/C

4 0
3 years ago
Help me please please?
Rainbow [258]

Answer: I looked it up and it says something about the waves traveling in a solid but I don’t know if that’s correct.

4 0
3 years ago
Read 2 more answers
A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.
Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

  1. Under the assumption that the can is a closed system, the conservation law applied to the system would be: E_{in}-E_{out}=E_{change}, where E_{in} is all energy entering the system, E_{out} is the total energy leaving the system and, E_{change} is the change of energy of the system.
  2. As the purpose is to kept the beverage can at constant temperature, the change of energy (E_{change}) would be 0.
  3. The energy  that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4) where \varepsilon is the emissivity of the surface, \sigma=5.67*10^{-8}\frac{W}{m^2K} known as the Stefan–Boltzmann constant, A_S is the total area of the exposed surface, T_S is the temperature of the surface in Kelvin, T_{\infty} is the environment temperature in Kelvin.
  4. For the can the surface area would be ta sum of the top and the sides. The area of the top would be A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2, the area of the sides would be A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2. Then the total area would be A_{total}=A_{top}+A_{sides}=0.01624m^2
  5. Then the radiation heat transferred to the can would be Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W.
  6. The can would lost heat evaporating water, in this case would be Q_{out}=\frac{dm}{dt}*h_{fg}, where \frac{dm}{dt} is the rate of mass of water evaporated and, h_{fg} is the heat of vaporization of the water (2257\frac{J}{g}).
  7. Then in the conservation balance: Q_{in}-Q_{out}=Q_{change}, it would be1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0.
  8. Recall that 1W=1\frac{J}{s}, then solving for \frac{dm}{dt}:\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}
5 0
3 years ago
The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This
erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

g = \frac{GM}{(d-R_{CM})^2}

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

R_{CM} = Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

g = \frac{GM}{(d-R_{CM})^2}

g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}

g = 3.4*10^{-5}m/s^2

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

\omega = \frac{2\pi}{T}

At the same time we have that centripetal acceleration is given as

a_c = \omega^2 r

Replacing

a_c = (\frac{2\pi}{T})^2 r

a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)

a_c = 3.34*10^{-5}m/s^2

3 0
3 years ago
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