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Olegator [25]
3 years ago
5

A parallel-plate capacitor has an area of 4.59 cm2, and the plates are separated by 1.28 mm with air between them. it stores a c

harge of 334 pc. what is the potential difference across the plates of the capacitor?
Physics
1 answer:
nalin [4]3 years ago
7 0
 <span>a) 
Capacitance = k x ε° x area / separation 
ε° = 8.854 10^-12 F/ m 
k = 2.4max 
average k = 0.78 / 1.27 * 2.4 +(1.27- 0.78) / 1.27 * 1 = 1.474 + 0.386 = 1.86 
(61.4 % separation k = 2.4 --- 38.6 % k = 1 air --- average k = 0.614 * 2.34 + 0.386 * 1 = 1.86 
area = 145 cm2 = 0.0145 m2 
separation = 1.27 cm 0.0127 m 

C = 1.86 * 8.854 10^-12 * 0.0145 / 0.0127 = 18.8 pF 
b) Q = C * V --- 18.8 * 83 = 1560.4 pC = 1.5604 nC 
c) E = V / d = 83 / 0.0127 = 6535.4 V/m </span>
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At what time was the person at a position of 0m?
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Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

  = (0 x 1) + 1/2 (0 x 1 x 1)

  = (0) + 1/2 (0)

  = 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

 = 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

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2 years ago
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</span>
8 0
3 years ago
100 POINTS FOR CORRECT ANSWER/EXPLANATION
Shalnov [3]

Answer:

6 N

Explanation:

Let's start with the small block m on top.  There are four forces:

Weight force mg pulling down, normal force N₁ pushing up, tension force T pulling right, and friction force N₁μ pushing left (opposing the direction of motion).

Now let's look at the large block M on bottom.  There are seven forces:

Normal force N₁ pushing down (opposite and equal from block m),

Friction force N₁μ pushing right (opposite and equal from block m),

Weight force Mg pulling down,

Tension force T pulling right,

Applied force F pulling left,

Normal force N₂ pushing up,

and friction force N₂μ pushing right (opposing the direction of motion).

So you've correctly identified the free body diagrams.

Now apply Newton's second law.  Sum of forces in the y direction for block m:

∑F = ma

N₁ − mg = 0

N₁ = mg

Sum of forces in the x direction:

∑F = ma

T − N₁μ = 0

T = N₁μ

T = mgμ

Sum of forces in the y direction for block M:

∑F = ma

-N₁ − Mg + N₂ = 0

N₂ = N₁ + Mg

N₂ = mg + Mg

Sum of forces in the x direction:

∑F = ma

N₁μ + T − F + N₂μ = 0

F = N₁μ + T + N₂μ

F = mgμ + mgμ + (mg + Mg)μ

F = gμ(3m + M)

Since M = 2m:

F = 5gμm

Plug in values:

F = 5 (10 m/s²) (0.400) (0.300 kg)

F = 6 N

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kodGreya [7K]

Answer:

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Explanation:

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based on the law of conservation of energy. how can we reasonably improve a machines ability to do work? A.move the machine to a
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The correct answer is

C. reduce the friction between its moving parts


In fact, by reducing the friction between the moving parts of the machine, it is possible to reduce the energy wasted due to this friction; therefore, more input energy is converted into useful work, and this will improve the efficiency of the machine.

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3 years ago
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