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Leno4ka [110]
3 years ago
13

What gravitational force does the moon produce

Physics
1 answer:
saul85 [17]3 years ago
6 0

Answer:

1.94\cdot 10^{20} N

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

The gravitational force is always attractive.

In this problem, we have:

m_1 = 5.98\cdot 10^{24}kg is the mass of the Earth

m_2 = 7.34\cdot 10^{22} kg is the mass of the Moon

r=3.88\cdot 10^8 m is the separation between the Earth and the Moon

Therefore, the gravitational force between them is

F=(6.67\cdot 10^{-11})\frac{(5.98\cdot 10^{24})(7.34\cdot 10^{22})}{(3.88\cdot 10^8)^2}=1.94\cdot 10^{20} N

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Answer:

The answer is already given... it's 31.25 cm

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As viewed from above in this picture, what direction will the current be in the coil of wire that will cause the loop to rotate
Gala2k [10]

Answer:

When viewed from above, the current in the coil should point towards the top-right corner of the picture.

Explanation:

The current in this coil have only two possible directions: clockwise or counter-clockwise. However, since the diagram shows the coil from above, not from a cross-section, just saying clockwise or counter-clockwise might be ambiguous. The statement that the current is directed towards the top-right corner of the picture is equivalent to saying that when viewed from the lower-right corner of this diagram, the current in the coil is moving clockwise.

Note that at the center of this picture, the current is parallel to the magnetic field- there will be no force on the coil at that position. On the other hand, (also when viewed from above,) at the top-right corner and the lower-left corner of the coil, the current in the coil will be perpendicular to the magnetic field. That's where the force on the coil will be the strongest.

With that in mind, apply the right-hand rule to find the direction of the force on the coil in each of the two possibilities.

Assume that when viewed from above, the current is flowing towards the top-right corner of the picture. Consider the wire near the top-right corner of this coil (as viewed above on this picture.) The current will be going into the picture into the magnetic field. By the right-hand rule, the current on the wire near that point should be pointing towards the bottom of this picture. (Point fingers on the right hand in the direction of the current I. Rotate the right hand such that when curling the fingers, they point in the direction of the magnetic field B. The direction of the right thumb should now point in the direction of the force on the wire F.)

Based on the same assumption, the current in the wires near the bottom left corner of this coil will be pointing out of the picture. By the right hand rule, the magnetic force on the coil in that region should be pointing towards the top of this picture. Combing these two forces, the coil would indeed be rotating around the center of this picture in the direction shown in the diagram.

It can also be shown that if the current points towards the bottom left corner of the picture when viewed from above, the coil will be rotating about the center of this picture in the opposite direction.

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Which configuration would produce an electric current? A) Rotate a coil of copper wire between two magnets. B) Connect a wire be
FrozenT [24]

the answer is (A) the movement of the magnet relative to the coil

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Read 2 more answers
An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
Veronika [31]

Answer:

s=5.79\ km

\theta=47^{\circ} east of south

Explanation:

Given:

  • distance of the person form the initial position, d'=8.4\ km
  • direction of the person from the initial position, 47^{\circ} north of east
  • distance supposed to travel form the initial position, d=5.3\ km
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<u>Now refer the schematic for visualization of situation:</u>

y=d'.\sin47^{\circ}-d

y=8.4\times \sin47-5.3 ...............(1)

x=d'.\cos47^{\circ}

x=8.4\times \cos47^{\circ} .................(2)

<u>Now the direction of the desired position with respect to south:</u>

\tan\theta=\frac{y}{x}

\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}

\theta=47^{\circ} east of south

<u>Now the distance from the current position to the desired position:</u>

s=\sqrt{x^2+y^2}

s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}

s=5.79\ km

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The potential difference across 3 Ohm resistor is 20V.

The resistors are connected in parallel which means all the three resistances have a fully potential difference of 20V.

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