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3 years ago

Why does the moons gravity affect the earth so much

Physics

2 answers:

Licemer1 [7]3 years ago

7 0

One main reason is, <span>The Moon's gravity pulls on Earth's oceans and distorts them, causing tides</span>

Ronch [10]3 years ago

6 0

The Moon's gravity pulls on Earth's oceans and distorts them, causing tides. The water on the side of Earth closest to the Moon experiences the biggest pull, and bulges outward. The water on the opposite side also bulges, and the two bulges follow the Moon's motion and Earth's rotation.

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Please help meeeeeeeeeeeeeeee

maxonik [38]

probabilityAnswer:

2/27

Explanation:

The elk can not be eaten so we remove that from the probablity

so we have x/18

songbird = 4/18

mice = 6/18

4/18*6/18 = 2/27

5 0

2 years ago

Does the KE of a car change more when it accelerates from 22 km/h to 32 km/h or when it accelerates from 32 km/h to 42 km/h

mote1985 [20]

Answer:

The change in kinetic energy (KE) of the car is more in the second case.

Explanation:

Let the mass of the car = m

initial velocity of the first case, u = 22 km/h = 6.11 m/s

final velocity of the first case, v = 32 km/h = 8.89 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

ΔK.E = ¹/₂m(8.89² - 6.11²)

= 20.85m J

initial velocity of the second case, u = 32 km/h = 8.89 m/s

final velocity of the second case, v = 42 km/h = 11.67 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

ΔK.E = ¹/₂m(11.67² - 8.89²)

= 28.58m J

The change in kinetic energy (KE) of the car is more in the second case.

6 0

3 years ago

A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c

viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = $71\times 10^{- 9} C$

Q' = + 42 nC = $42\times 10^{- 9} C$

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

$\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E} = 708.03 N/C$

Now,

Electric field at the mid-point due to charge Q' is given by:

$\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E'} = 418.84 N/C$

Now,

The net Electric field is given by:

$\vec{E_{net}} = \vec{E} - \vec{E'}$

$\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C$

5 0

3 years ago

It has made us think we have to live perfect lives?

Citrus2011 [14]

Answer: Is this a question? I dont understand.

Explanation:

3 0

3 years ago

What is the difference between a circuit and a current?

dimulka [17.4K]

Answer:

a circuit is a complete loop that carries a current in the form of electrons from negative to positive

Explanation:

The difference between current and circuit is that a circuit is a complete loop that carries a current in the form of electrons from negative to positive .

In a circuit, there are difference elements or components such as the battery, wire, resistor. The goal of a circuit arrangement is to completely carry current from one end to another.

The current is the quantity of charge that flows within the circuit per unit of time.

So, the battery supplies the electromotive force to move the current round the circuit.

3 0

3 years ago