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Klio2033 [76]
3 years ago
13

Describe two possibilities of what will happen to the universe in the future.

Physics
1 answer:
shutvik [7]3 years ago
4 0
In our current view of the Universe there are 2 possible futures. One possiblity is that the Universe will come to an end in the opposite of a Big Bang called The Big Crunch. The other possibility is that we live in an Eternal Universe that will never come to end.

i hope this helps :)
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Which two, or more, of the following actions would increase the energy stored in a parallel plate capacitor when a constant pote
MArishka [77]

1. increase the area of the plates

4. decrease the separation between the plates

5. insert a dielectric between the plates

Explanation:

The energy stored in a capacitor is given by

U=\frac{1}{2}CV^2

where

C is the capacitance

V is the potential difference across the capacitor

For a parallel-plate capacitor, the capacitance is given by

C=\frac{k \epsilon_0 A}{d}

where

k is the dielectric constant of the material

\epsilon_0 is the vacuum permittivity

A is the area of the plates of the capacitor

d is the separation between the plates

So we can rewrite the energy stored in the capacitor as

U=\frac{k \epsilon_0 A V^2}{2d}

Here the potential difference is kept constant, so the energy depends only on the dielectric constant of the medium, the area and on the distance between the plates. In particular:

- The energy is directly proportional to the area, so as the area increases, the energy will increase

- The energy is inversely proportional to the distance, so as the distance decreases, the energy will increase

- The energy increases if the value of k increases (that is, if a dielectric is put between the plates)

It follows that the correct options to increase the energy are:

1. increase the area of the plates

4. decrease the separation between the plates

5. insert a dielectric between the plates

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

#LearnwithBrainly

6 0
4 years ago
How much work did the movers do (horizontally) pushing a 41.0-kg crate 10.6 m across a rough floor without acceleration, if the
VLD [36.1K]

Answer:

The required work done is 2555.448~J

Explanation:

Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if '\mu_{k}' be the coefficient of friction, then

f = \mu_{k} \times N = \mu_{k} \times Mg

where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.

Since the application of force by the movers does not create any acceleration to the block, we can write

F = f = \mu_{k} \times M \times g = 0.6 \times 41~Kg~ \times 9.8~m~s^{-2} = 241.08~N

So the work done (W) in moving the crate by a distance s = 10.6 m is

W = F \times s = 241.08~N \times 10.6~m = 2555.448 J

5 0
3 years ago
If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattenti
Ann [662]
We know that speed equals distance between time. Therefore to find the distance we have that d = V * t. Substituting the values d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km.Therefore during this inattentive period traveled a distance of 0.08Km
8 0
3 years ago
Using a rope that will snap if the tension in it exceeds 356 N, you need to lower a bundle of old roofing material weighing 478
klasskru [66]

Answer:

a) 2.5 m/s²

b) 6.12 m/s

Explanation:

Tension of rope = T = 356N

Weight of material = W = 478 N

Distance from the ground = s = 7.5 m

Acceleration due to gravity = g = 9.81 m/s²

Mass of material = m = 478/9.81 = 48.72

Final velocity before the bundle hits the ground = v

Initial velocity = u = 0

Acceleration experienced by the material when being lowered = a

a) W-T = ma

⇒478-356 = 48.72×a

\Rightarrow \frac{122}{48.72} = a

⇒a = 2.5 m/s²

∴ Acceleration achieved by the material is 2.5 m/s²

b) v²-u² = 2as

⇒v²-0 = 2×2.5×7.5

⇒v² = 37.5

⇒v = 6.12 m/s

∴ Velocity of the material before hitting the ground is 6.12 m/s

5 0
3 years ago
A 10.0-g bullet moving at 300 m/s is fired into a 1.00-kg block at rest. the bullet emerges (the bullet does not get embedded in
sesenic [268]
Momentum before the hit:
p = mv = 0.01 * 300 + 1 * 0 

Momentum after the hit:
p = 0.01 * 150 + 1 * v

Momentum is conserved:
0.01 * 300 = 0.01 * 150 + v 
3 = 1.5 + v
v = 1.5

The velocity of the block after the collision is 1.5 m/s.
7 0
4 years ago
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