Ask question

Ask question

All categories

Montano1993 [528]

4 years ago

10

To get an idea how big a farad is, suppose you want to make a 1-F air-filled parallel-plate capacitor for a circuit you are buil

ding. To make it a reasonable size, suppose you limit the plate area to What would the gap have to be between the plates? Is this practically achievable?

1 answer:

Triss [41]4 years ago

4 0

Answer:

Gap between the plates $8.85\times 10^{- 16}\ m$
No, practically not achievable

Solution:

As per the question:

Capacitance, C = 1 F

Area of the plate of the capacitor, A = $1\ cm^{2} = 1\times 10^{- 4}\ m^{2}$

Now,

To calculate the distance, D between the plates of a parallel plate capacitor:

$C = \frac{\epsilon_{o}A}{D}$

Thus

$D = \frac{\epsilon_{o}A}{C}$

where

$\epsilon_{o}$ = permittivity of free space

Now,

$D = \frac{8.85\times 10^{- 12}\times 1\times 10^{- 4}}{1}$

$D = 8.85\times 10^{- 16}\ m$

This distance much smaller and is practically not possible

You might be interested in

An astronaut is out in space with an extremely precise timing device measuring the speed of various fast‑moving objects. A laser

Leviafan [203]

Answer and Explanation:

with reference to Einstein's theory of special relativity, the speed of an electromagnetic radiation, here, laser will not change in any inertial frame or remains same irrespective of any change in inertial frame.

Therefore, the speed of light measured in both the cases, i.e., in astronaut's reference frame and spaceship's reference frame will be equal to the speed of light in vacuum, i.e., $3\times 10^{8} m/s$ .

The laser gun's speed in astronaut's reference frame is the same as the speed of the spaceship as it mounted on it, i.e., the speed of the laser gun is 200 million m/s.

The laser gun's speed measured in spaceship's reference frame will be zero, as it is mounted on the spaceship and is stationary in the spaceship's reference frame.

5 0

3 years ago

What is the measure of how much a material resists the formation of an electric field?

shusha [124]

The answer is c capacitance

5 0

4 years ago

A high school physics student with a mass of 68.18 KG is sitting in a seat reading this question the magnitude of the force with

love history [14]

The answer to the question is C

3 0

3 years ago

A 99.5 N grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 N. The acceler

Romashka [77]

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

$W=K_f - K_i$

Since the cart was initially at rest, $K_i = 0$ , so

$W=K_f = \frac{1}{2}mv^2$ (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, $F_g = 99.5 N$ :

$m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg$

So solving eq.(1) for v, we find the final speed of the cart:

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s$

2) $2.51\cdot 10^7 J$

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

$F=4.28 \cdot 10^5 N$

$d=586 m$

So the work done is

$W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J$

3) $2.51\cdot 10^7 J$

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

$W=\Delta K = K_f - K_i$

where

W is the work done

$\Delta K$ is the change in kinetic energy

Therefore, the change in kinetic energy is

$\Delta K = W = 2.51\cdot 10^7 J$

4) 37.2 m/s

According to the work-energy theorem,

$W=\Delta K = K_f - K_i$

where

$K_f$ is the final kinetic energy of the train

$K_i = 0$ is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

$W=K_f = \frac{1}{2}mv^2$

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s$

7 0

4 years ago

Larger animals use proportionately less energy than smaller animals; that is, it takes less energy per kg to power an elephant t

marin [14]

Answer:

Part a)

$Q = 952 cal/day$

Part b)

$P = 46 Watt$

Part c)

$\Delta P = 54 W$

Explanation:

As we know that 5000 kg African elephant requires 70,000 Cal for basic needs per day

so we will have

$m = 5000 kg$

$Q = 70,000 Cal$

so we have energy required per kg

$E = \frac{70,000}{5000}$

$E = 14 cal/kg$

Part a)

now we know that per kg the energy required will be same

so we have mass of the human is 68 kg

so energy required per day is given as

$Q = 68 \times 14$

$Q = 952 cal/day$

Part b)

Resting power is the rate of energy in Joule required per sec

so it is given as

$P = \frac{952 \times 4186}{24\times 3600}$

$P = 46 Watt$

Part c)

resting power given in the book is

$P' = 100 W$

so this is less than the power given

$\Delta P = 100 - 46$

$\Delta P = 54 W$

6 0

3 years ago