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3 years ago

What would happen to the moon if earth stopped exerting the force of gravity on it?

2 answers:

6 0

There are two equal forces of gravity between the Earth and the Moon.
One force pulls the Moon toward the Earth.
The other force pulls the Earth toward the Moon.

If only this gravity suddenly switched off, then the moon would
continue to orbit the Sun, very much as it does now.

If ALL gravity suddenly switched off, then . . .

-- the Moon would stop orbiting the Earth and would sail away, in
a straight line and at the speed it had when gravity disappeared;

-- the Earth would stop orbiting the Sun and would sail away, in
a straight line and at the speed it had when gravity disappeared;

-- all the gases surrounding the Earth ... which we call "air" ... would
start drifting away, and expanding into a giant cloud of gas, and stop
being an atmosphere;

-- the Sun would completely fall apart, expand into a giant cloud of gas,
and stop being a star.

larisa86 [58]3 years ago

3 0

The moon will stop orbiting and will fly away and never stop running

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Which wave, A or B, has higher energy?

eduard

Answer:

Wave A.

Explanation:

The energy of a wave is directly proportional to the square of the amplitude.

If a wave has higher amplitude, it will have more energy. On the other hand, a wave having lower amplitude, it will have less eenergy.

In this case, we need to tell which wave has higher energy. Hence, the correct option is A because it has a higher amplitude.

8 0

3 years ago

Read 2 more answers

Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th

Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

$\theta=30^{\circ}$

So, the ball's horizontal velocity is

$v_x = u cos \theta = (15)(cos 30)=13.0 m/s$

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

$y=u_yt+\frac{1}{2}at^2$

where

$u_y = u sin \theta = (15)(sin 30) = 7.5 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

The ball is at a position of y = 2 m above the ground when:

$2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0$

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

$d=v_x t =(13.0)(1.19)=15.5 m$

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

$v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s$

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

$d=98 m$

while the horizontal velocity of the ball is

$v_x = u cos \theta = (34)(cos 30)=29.4 m/s$

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

$t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s$

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

$y= h + u_y t + \frac{1}{2}at^2$

where

h = 1.2 m is the initial height

$u_y = u sin \theta = (34)(sin 30)=17.0 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Substituting t = 3.33 s,

$y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m$

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

$u_y' = 7 m/s$

So its position of the glove at time t' is

$y'= h' + u_y' t' + \frac{1}{2}at'^2$

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

$3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0$

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

$t'' = t -t'$

So we have two solutions:

$t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s$

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

A 70 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 590 m/s . Part A Part complete Wha

Kisachek [45]

Answer:

The recoil velocity is 0.354 m/s.

Explanation:

Given that,

Mass of hunter = 70 kg

Mass of bullet = 42 g = 0.042 kg

Speed of bullet = 590 m/s

We need to calculate the recoil speed of hunter

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Where, $m_{1}$ = mass of hunter

$m_{2}$ = mass of bullet

u = initial velocity

v = recoil velocity

Put the value in the equation

$0+0=70\times v_{1}+0.042\times590$

$v_{1}=-\dfrac{0.042\times590}{70}$

$v=-0.354\ m/s$

Hence, The recoil velocity is 0.354 m/s.

8 0

3 years ago

How does adding more of a substance affect it's density?

fomenos

I’m going to use molasses as an example of a substance.

The mass and volume both change when changing the amount of molasses.
However, the density does not change. This is because the mass and volume increase at the same rate/proportion!

Even though there is more molasses (mass) in test tube A, the molasses also takes up more space (volume). Therefore, the spacing between those tiny particles that make up the molasses is constant (does not change).

The size or amount of a material/substance does not affect its density.

5 0

4 years ago

Read 2 more answers

The kitchen in your house measures 10 ft by 10 ft. You need to have 50 footcandles of illumination in the room. How many 60 watt

tekilochka [14]

Answer:

<u>6 bulbs</u> are needed to illuminate the room.

Explanation:

Given:

Measurement of kitchen (A) = 10 ft by 10 ft = 100 sq. ft

Number of footcandles (n) = 50

Lumens emitted by 1 bulb = 834

Number of bulbs (N) = ?

We are also given,

1 foot candle = 1 lumen/sq. ft

So, 50 foot candles = 50 lumens/sq. ft

Now, for an area of 1 sq. ft 50 lumens are emitted.

So, for an area of 100 sq. ft, lumens emitted = 50 × 100 = 5000 lumens

Now, one bulb emits = 834 lumes

Therefore, number of bulbs required for emitting 5000 lumens is given as:

$N=\frac{Number\ of\ lumens}{Number\ of\ lumens\ by\ 1\ bulb}\\\\N=\frac{5000}{834}=5.995\approx6$

So, 6 bulbs are needed to illuminate the room.

7 0

3 years ago