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Anna11 [10]
3 years ago
5

Would time travel mean that parallel universes would be created off our linear time sequence to ensure the universe doesn't coll

apse itself because of significant events being drastically changed in the way it shapes the universe?
Physics
2 answers:
MatroZZZ [7]3 years ago
4 0
In theroy, yes. But time travel has yet to happen and thus, we dont know. I would assume that that if an alternate universe were NOT created, it would cause a rip in the fabic of time creating a worm hole and allowing all entitys to pass through simultaneously,and that would cause great stress on it, possibly causing it to rip large enough to encase the whole universe and it would implode.
Luden [163]3 years ago
3 0

My impression of that question is that you've got enough
made-up concepts and buzzwords there to get started on
a great science-fiction story.  You should pick up from there
and write a scene about what time travel does to our universe. 

The other answer, written by "Brainlychamp", also has some great
techy-sounding phrases, like "rip in the fabric of time", that you could
use if he hasn't already copyrighted them.

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List five situations in which a lesser amount of friction would be beneficial. Overall, would a high or low coefficient of frict
Natalka [10]
Low coefficient of friction

1. flying a plane (friction between air and plane)
2. ice skating (friction between ice and skate blade)
3. swimming (water & skin)
4. rowing a boat (water and boat)

6 0
3 years ago
Read 2 more answers
PLZZZZZ ANSWER Which example describes a nonrenewable resource?
Zepler [3.9K]

Answer:

I believe that the answer is D. There are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity.

Explanation:

Solar panels use the sun, and that is renewable.

The power plant uses tides and waves, they are renewable.

Windmills use wind, that is renewable.

So, the answer is D.

7 0
2 years ago
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A student throws a 120 g snowball at 7.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of
mr Goodwill [35]

Answer:

The magnitude of the average force on the wall during the collision is 6 N.

Explanation:

Given;

mass of snowball, m = 120 g = 0.12 kg

velocity of the snowball, v = 7.5 m/s

duration of the collision between the snowball and the wall, t = 0.15 s

Magnitude of the average force can be calculated by applying Newton's second law of motion;

F = ma

where;

a is acceleration = v / t

a = 7.5 / 0.15

a = 50 m/s²

F = ma

F = 0.12 x 50

F = 6 N

Therefore, the magnitude of the average force on the wall during the collision is 6 N.

4 0
3 years ago
While driving on the highway, you have a tire blowout. after you have slowed down and regained control, you should?
Svetllana [295]
When a person has experienced a tire blow out, there are steps to follow in order to prevent harm to others and self. After following the required and helpful steps after a tire blow out, slowed down and regain control, it is best that the person should head to a stop road or at the safe side of the lane where they won't be a bother to the road or to people driving in the high way. After pulling over to the side, it is advisable to turn on the emergency flashers of the car. This will set as a signal that you are in need of help or had gone through a problem.
4 0
3 years ago
A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on
SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

\sum F = ma

the resultant of the forces must be zero: \sum F = 0 (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

F, the push applied by the worker

F_f=-\mu mg, the force of friction, with \mu=0.25 being the coefficient of friction, m=30.0 kg being the mass of the crate, and g=9.8 m/s^2. The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

W=Fd cos \theta

where

F = 73.5 N is the force

d = 4.5 m is the displacement

\theta=0^{\circ} is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

W=F_f d cos \theta

where

F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N is the magnitude of the force of friction

d = 4.5 m is the displacement

\theta=180^{\circ} is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

W=F_f d cos \theta

We notice that both gravity and normal force are perpendicular to the displacement: therefore, \theta=90^{circ}, and so

cos \theta=0

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0
2 years ago
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