Which exerts more force, the Earth pulling on the moon or the moon pulling on the Earth? Explain.

Answers

Answer 1

Answer:

Answer: the earth

Explanation: Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth. Over a very long time, the moon's rotations created fiction with the Earth's tugging back, until the moon's orbit and rotational locked with Earth.

and that's why the earth pulls the moon

Answer 2

Answer:

Final answer:

The Earth pulling on the moon and the moon pulling on the Earth exert the same amount of force on each other due to Newton's third law of motion.

Explanation:

In terms of force, the Earth pulling on the Moon and the Moon pulling on the Earth exert the same amount of force on each other. This is because of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. So, while the Earth's gravitational force pulls the Moon towards it, the Moon's gravitational force also pulls the Earth towards it with an equal amount of force.

Newton's third law of motion states that for every action, there is an equal and opposite reaction. In the context of the gravitational interaction between the Earth and the Moon, the forces they exert on each other are equal in magnitude and opposite in direction.

The Earth pulls on the Moon with a gravitational force, and, according to Newton's third law, the Moon simultaneously pulls on the Earth with an equal gravitational force. These forces are sometimes referred to as "action and reaction pairs." The force that the Earth exerts on the Moon is often called the gravitational attraction of the Earth on the Moon, and vice versa.

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If a photon has a frequency of 5.20 x 10^14 hertz, what is the energy of the photon ? Given : Planck's constant is 6.63 x 10^-34 joule-seconds.

Answers

For this you would use planck's equation.

E = hv, where v = the frequency and h = planck's constant.

So E = 5.20 x10^14 x 6.63 x 10^-34
= 3.45 x 10^-19 Joules

The distance in the x direction between two control points on a vertical aerial photograph is 4.5". If the distance between these same two points is 3.6" on another photograph having a scale of 1:24,000, determine the scale of the first vertical aerial photograph. Of the focal length of the camera is 6"and the average elevation at these points is 100 ft, determine the flying height from which each photograph was taken

Answers

Answer:

Use proportions to find the scale of the first photo, then use that scale and other given information to fill in the equation

S=f/(H-h)

Where:

S = scale of the photo

f = focal length of the camera (in feet)

H = flying height

h = average elevation

Look at the Kunoichi's of Naruto but as a Gang. Who do you think looks the baddest out of the group(but in a good way)?

My opinion is Hinata...just saying

Answers

Answer:

hinata for sure

Explanation:

seems reasonable

An implanted pacemaker supplies the heart with 72 pulses per minute, each pulse providing 6.0 V for 0.65 ms. The resistance of the heart muscle between the pacemaker’s electrodes is 550 Ω. Find (a) the current that flows during a pulse, (b) the energy delivered in one pulse, and (c) the average power supplied by the pacemaker.

Answers

Answer:

a) Current = 11 mA

b) Energy = 66 mJ

c) Power = 101.54 W

Explanation:

a) Voltage, V = IR

Voltage, V = 6 V, Resistance, R = 550 Ω

Current, I $=(6)/(550)=0.011A=11mA$

b) Energy = Current x Voltage = 6 x 0.011 = 0.066 J = 66 mJ

c) $\texttt{Power=}(Energy)/(Time)=(0.066)/(0.65* 10^(-3))=101.54W$

A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very far from the wire. What is the ratio Ff/Fi of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk?

Answers

The ratio of the electric force on the proton after the wire segment is shrunk to three times its original length to the force before the segment was shrunk is 3.

The electric force between a point charge and a segment of wire with a distributed charge is given by Coulomb's law.

The formula for the electric force on a point charge q due to a segment of wire with charge Q distributed along its length L is:

$F=(k.q.Q)/(L)$

where:

F is the electric force on the point charge,

k is Coulomb's constant ( 8.988 × 1 0⁹ Nm²/ C²),

q is the charge of the point charge,

Q is the charge distributed along the wire segment, and

L is the length of the wire segment.

When the wire segment is shrunk to one-third of its original length, the new length becomes 1/3 L.

The chargedistribution remains the same, only the length changes.

So, the new electric force $F_f$ on the proton after the segment is shrunk becomes:

$F_f=(k.q.Q)/((1)/(3)L)$

The original electric force $F_i$ on the proton before the segment was shrunk is:

$F_i = (k.q.Q)/(L)$

let's find the ratio $(F_f)/(F_i)$ :

$(F_f)/(F_i)=((k.q.Q)/((1)/(3)L))/((k.q.Q)/(L))$

$(F_f)/(F_i)=3$

Hence, the ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk is 3.

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Final answer:

The ratio of the electric force on the proton after the wire segment is shrunk is equal to the ratio of their charges.

Explanation:

The ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk can be found using Coulomb's law. Coulomb's law states that the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In this case, the charges involved are the charge of the wire segment and the charge of the proton. Since the wire segment contains 10 nC of charge, we can consider it as one of the charged objects. The proton is very far from the wire, so we can assume that the distance between them remains the same before and after the wire segment is shrunk. Therefore, the ratio of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk is equal to the ratio of their charges.

Let's assume that the initial force on the proton is Fi and the final force on the proton is Ff. Using the given information, we have:

Fi = k(q1 * q2) / r^2

where k is the electrostatic constant, q1 and q2 are the charges of the wire segment and the proton respectively, and r is the distance between them.

After the wire segment is shrunk to one-third of its original length, the charge of the wire segment remains the same and the distance between the wire segment and the proton also remains the same. Therefore, the ratio Ff/Fi can be calculated as:

Ff/Fi = (q1 * q2) / (q1 * q2) = 1

When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to make this force twice as strong, the distance would have to be changed toA) √2d
B) d/√2
C) d/4
D) 2d
E) d/2