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3 years ago

What might Earth be like if it had never been hit by the theoretical protoplanet Orpheus?

Physics

1 answer:

Dafna1 [17]3 years ago

7 0

Answer:

If Earth hadn't been hit by Orpheus, it would be covered by ocean, with perhaps a few mountaintops emerging through the water. There would be no humans, but there could be other forms of life. Earth would rotate rapidly, as the moon would not be present to produce the tidal friction that slows Earth's rotation today

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A Tennis ball falls from a height 40m above the ground the ball rebounds

worty [1.4K]

If the ball is dropped with no initial velocity, then its velocity v at time t before it hits the ground is

v = -g t

where g = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height y is

y = 40 m - 1/2 g t²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 g t²

==> t = √(80/g) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

v = -g (√(80/g) s)

==> v = -√(80g) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

y = (14 m/s) t - 1/2 g t²

Solve y = 0 for t to see how long it's airborne during this bounce:

0 = (14 m/s) t - 1/2 g t²

0 = t (14 m/s - 1/2 g t)

==> t = 28/g s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

y = (14 m/s) (5 s - 2.86 s) - 1/2 g (5 s - 2.86 s)²

==> (i) y ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But y will converge to 0 as t gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

y = (3.5 m/s) t - 1/2 g t²

Solve y = 0 for t :

0 = (3.5 m/s) t - 1/2 g t²

0 = t (3.5 m/s - 1/2 g t)

==> (iii) t = 7/g m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time t after 5.72 s (but before reaching the ground again) would be

v = 7 m/s - g t

At 6 s, the ball has velocity

(iv) v = 7 m/s - g (6 s - 5.72 s) ≈ 4.26 m/s

4 0

3 years ago

A thermometer is placed in water in order to measure the water’s temperature. What would cause the liquid in the thermometer to

katovenus [111]

The correct answer should be c.The kinetic energy of the water molecules decreases.

If the temperature drops that means that the molecules are coming together. If the temperature rises then it means that the molecules are spreading. If the kinetic energy falls down that means that they are slower which means that they are cooler.

4 0

3 years ago

Read 2 more answers

A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base

umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388 m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388 m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0

3 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

Lorentz transformation

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$