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miv72 [106K]
2 years ago
6

How does the configuration of the electric field occur between a "parallel plate" setup in a lab?What is the effect of conductor

s (rings and the triangle plates) on the electric field and equipotential lines?
Physics
1 answer:
sleet_krkn [62]2 years ago
8 0

Explanation:

The magnitude of the electric field between the plates is given by

E = -ΔV/d

minus sign indicates Potential decreases in the direction of electric field

where

ΔV is the potential difference between the plates

D is the distance between the plates.

The work done when carrying an electrical charge on an equipotential surface between one position to the other is zero W= q(V-V)=0 The electric field lines of force are always perpendicular to an equipotential surface.  That conductor in an equipotential surface as direction E is at right angles to an eauipotential surface The intensity of the electric field along an equipotential surface is always zero. Equipotential surfaces never collide with each other as this would mean that at that point, there are two alternative values that are not true.

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A hammer strikes one end of a thick iron rail of length 5.60 m. A microphone located at the opposite end of the rail detects two
never [62]

Answer:

A) The wave that travels through the rail reaches the microphone first.

B) separation in time between the arrivals of the two pulses is 0.01539 seconds.

Explanation:

Detailed explanation and calculation is shown in the image below

4 0
3 years ago
A block with mass m 2.00 kg is placed against a spring on a frictionless incline with angle 30 degrees (Fig. B-43). (The block i
guajiro [1.7K]

Answer:

Explanation:

a )

The stored elastic energy of compressed spring

= 1 / 2 k X²

= .5 x 19.6 x (.20)²

= .392 J

b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .

c ) Let h be the distance along the incline which the block ascends.

vertical height attained ( H ) =h sin30

= .5 h

elastic potential energy = gravitational energy

.392 = mg H

.392 = 2 x 9.8 x .5 h

h = .04 m

4 cm .

=

7 0
3 years ago
Pweese help luvs qwq<br> pweese look at the image?
hjlf

Answer:

12 m/s

Explanation:

divide distance over time

72/6 = 12

6 0
2 years ago
Suppose that you are standing on a train accelerating at 0.39g. What minimum coefficient of static friction must exist between y
timurjin [86]

Answer:

0.39

Explanation:

In order not to slide, you must have exactly the same acceleration of the train:

a=0.39 g

where

g = 9.81 m/s^2

There is only one force acting on you: the static frictional force that "pulls" you forward, and it is given by

F_s = \mu_s mg

According to Newton's second law, the net force acting on you (so, the frictional force) must be equal to your mass times the acceleration, so we have

F= ma = \mu_s mg

from which we find

\mu_s = \frac{a}{g}=\frac{0.39 g}{g}=0.39

so, the minimum coefficient of static friction must be 0.39.

5 0
3 years ago
2 / 3
kiruha [24]

Answer:

Explanation:

   

3 0
3 years ago
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