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Alecsey [184]
2 years ago
11

A plate in a parallel-plate capacitor has an area of 0.03 m2 and is 0.5 × 10–3 m from the other plate. The space between the pla

tes is filled with a dielectric that has a dielectric constant of 6.7. Determine the capacitance of the capacitor.
Recall that ε0 = 8.85 × 10–12
Physics
1 answer:
Tems11 [23]2 years ago
4 0

Answer:

A. 4 × 10–9 F

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A ball is thrown vertically upward with a speed of 27.9 m/s from a height of 2.0 m. How long does it take to reach its highest p
Bas_tet [7]

Answer:

2.84403 seconds

2.91483 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-27.9}{-9.81}\\\Rightarrow t=2.84403\ s

It takes 2.84403 seconds to reach the highest point

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-27.9^2}{2\times -9.81}\\\Rightarrow s=39.67431 m

The ball will travel 39.67431+2 = 41.67431 m while going down to the ground

s=ut+\frac{1}{2}at^2\\\Rightarrow 41.71479=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{41.67431\times 2}{9.81}}\\\Rightarrow t=2.91483\ s

The ball takes 2.91483 seconds to hit the ground after it reaches its highest point.

6 0
2 years ago
How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?
charle [14.2K]
<span>The change in internal energy is only gravitional PE because the tube is being drug up at a constant speed. Since it is at a constant speed, the change in KE is 0. Change in PE = m*g*h = 78 kg * 10 m/s^2 * 30 m = 23400 J Work done on the system is from the force Work = force * distance = 350 N * 120 m = 42000 J So, work added 42000 J to the system, but the rider's energy only increased 23400 J. Therefore, friction took up the difference. Friction is where the thermal energy comes from Q = 42000 J - 23400 J = 18600 J. Therfore, friction generated 18600 J of heat to the surroundings.</span>
7 0
3 years ago
an experiment is set up to measure the effect of a gasoline additive on fuel consumption rate. what would be the control in this
Tamiku [17]
The control setup in this experiment would be one tank that does not contain any of the additives. Since the tanks with the gasoline additives would need to be compared with a tank that is not affected by the results of these additives.
4 0
2 years ago
A 600g toy train completes 10 laps of its circular track in 1 min 20s. If the radius of the track is 1.2 m, Find the centripetal
Lynna [10]

Wow !  This will take more than one step, and we'll need to be careful
not to trip over our shoe laces while we're stepping through the problem.

The centripetal acceleration of any object moving in a circle is

                          (speed-squared)  /  (radius of the circle)  .

Notice that we won't need to use the mass of the train.

We know the radius of the track.  We don't know the trains speed yet,
but we do have enough information to figure it out.  That's what we
need to do first.

Speed  =  (distance traveled) / (time to travel the distance).

Distance = 10 laps of the track.   Well how far is that ? ? ?

1 lap = circumference of the track = (2π) x (radius) =  2.4π  meters

10 laps =  24π  meters.

Time = 1 minute 20 seconds  =  80 seconds

The trains speed is  (distance) / (time)

                               =  (24π meters) / (80 seconds)

                               =        0.3 π  meters/second .

NOW ... finally, we're ready to find the centripetal acceleration.

                                 <span> (speed)²  /  (radius)

                           =    (0.3π m/s)²  /  (1.2 meters)

                           =    (0.09π m²/s²)  /  (1.2 meters)

                           =    (0.09π  /  1.2)   m/s²

                           =          0.236  m/s²  .        (rounded)

If there's another part of the problem that wants you to find
the centripetal FORCE ...

Well,       Force = (mass) · (acceleration) .

We know the mass, and we ( I ) just figured out the acceleration,
so you'll have no trouble calculating the centripetal force.       </span>
4 0
2 years ago
Select all that apply. There MIGHT be more than one.
Stolb23 [73]
I believe the answer to your question is “Lithosphere plate boundaries”

The planet Earth is covered by a layer formed by land and rocks called the earth's crust or lithosphere. This crust is not smooth and uniform, but rather irregular and composed of tectonic plates, also called lithosphere plates. These plates are not fixed as they are under the magma (high temperature molten rock).

Hope this helps!:)
4 0
2 years ago
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