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3 years ago

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

Physics

1 answer:

prohojiy [21]3 years ago

6 0

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

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Water flows from the bottom of a storage tank at a rate of r(t) = 300 − 6t liters per minute, where 0 ≤ t ≤ 50. Find the amount

SSSSS [86.1K]

r(t) models the water flow rate, so the total amount of water that has flowed out of the tank can be calculated by integrating r(t) with respect to time t on the interval t = [0, 35]min

∫r(t)dt, t = [0, 35]

= ∫(300-6t)dt, t = [0, 35]

= 300t-3t², t = [0, 35]

= 300(35) - 3(35)² - 300(0) + 3(0)²

= 6825 liters

7 0

3 years ago

How much work is it to lift a 6.35 kg Simon 2.14 meters to the top of his cat tree

Sloan [31]

Answer:

135.89J

Explanation:

weight = mass × gravitational field strength

= 6.35kg × 10N/kg

= 63.5N

work done = force × distance

= 63.5N × 2.14m

= 135.89J

8 0

3 years ago

What is one of the earliest ways that ancient peoples produced light?

VARVARA [1.3K]

Answer:

Candles

Explanation:

The answer is the first option or "candles." In ancient times the people would lit candles and place them in difference places in their homes, temples, etc... to produce light. It's not option B because light emitting diodes are used to light up toys and such which weren't invented in ancient times. It's not option C because fluorescent bulbs are long light bulbs used to light up huge spaces without as many light bulbs which were again not invented at the time, and it's also not option D because incandescent bulbs are you usual light bulbs which weren't invented at the time.

Hope this helps.

4 0

3 years ago

A toy cannon uses a spring to project a 5.38-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force

yawa3891 [41]

(a) 1.43 m/s

We can solve this problem by using the law of conservation of energy.

The initial total energy stored in the spring-mass system is

$E=U=\frac{1}{2}kx^2$

where

k = 7.91 N/m is the spring constant

$x=5.08 cm = 0.0508 m$

Substituting,

$E=\frac{1}{2}(7.91)(0.0508)^2=0.0102 J$

The final kinetic energy of the ball is equal to the energy released by the spring + the work done by friction:

$E+W_f=K$

where

$K_f=\frac{1}{2}mv^2$ is the kinetic energy of the ball, with

$m=5.38 g = 5.38\cdot 10^{-3} kg$ being the mass of the ball

v being the final speed

$W_f = -F_f d$ is the work done by friction (which is negative since the force of friction is opposite to the motion), where

$F_f = 0.0323 N$ is force of friction

d = 14.5 cm = 0.145 m is the displacement

Substituting,

$W_f = -(0.0323)(0.145)=-4.68\cdot 10^{-3} J$

So, the kinetic energy of the ball as it leaves the cannon is

$K_f = E+W_f = 0.0102 - 4.68\cdot 10^{-3}=0.00552 J$

And so the final speed is

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00552)}{0.00538}}=1.43 m/s$

(b) +5.08 cm

The speed of the ball is maximum at the instant when all the elastic potential energy stored in the spring has been released: in fact, after that moment, the spring does no longer release any more energy, so the kinetic energy of the ball from that moment will start to decrease, due to the effect of the work done by friction.

The elastic potential energy of the spring is

$U=\frac{1}{2}kx^2$

And this has all been released when it becomes zero, so when x = 0 (equilibrium position of the spring). However, the spring was initially compressed by 5.08 cm, so the ball has maximum speed when

x = +5.08 cm

with respect to the initial point.

(c) 1.78 m/s

The maximum speed is the speed of the ball at the moment when the kinetic energy is maximum, i.e. when all the elastic potential energy has been released.

As we calculated in part (a), the total energy released by the spring is

E = 0.0102 J

The work done by friction here is just the work done to cover the distance of

d = 5.08 cm = 0.0508 m

Therefore

$W_f = -(0.0323)(0.0508)=-1.64\cdot 10^{-3} J$

So, the kinetic energy of the ball at the point of maximum speed is

$K_f = E+W_f = 0.0102 - 1.64\cdot 10^{-3}=0.00856 J$

And so the final speed is

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00856)}{0.00538}}=1.78 m/s$

7 0

3 years ago

In a game of ice hockey, you use a hockey stick to hit a puck of mass 0.16 kg that slides on essentially frictionless ice. Durin

Grace [21]

Answer:

12N

Explanation:

given- mass, acelation

Fnet=ma= .16kg*75m/s2

Fnet=12 N only force no friction given.

8 0

3 years ago

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

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