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iVinArrow [24]
3 years ago
8

A dielectric material such as paper is placed between the plates of a capacitor holding a fixed charge. What happens to the elec

tric field between the plates
Physics
1 answer:
MAVERICK [17]3 years ago
5 0

Answer:

Majorly the electric field is reduced among other effect listed in the explanation

Explanation:

In capacitors the presence of di-electric materials

1. decreases the electric fields

2. increases the capacitance of the capacitors.

3. decreases the voltage hence limiting the flow of electric current.

 The di-electric material serves as an insulator between the metal plates of the capacitors

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Choose the correct statement about the acceleration of the car.
brilliants [131]

Answer:

Whats the question/word problem or where is the graph (if included) representing this problem?

4 0
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Each roller under a conveyor belt has a radius of 0.5 meters. the rollers turn at a rate of 30 revolutions per minute. what is t
svp [43]
Distance in a minute=<span>0.5 times 30=15 meters
distance in a second</span><span>=15 divided by 60=0.25 meters per second
hope it helps</span>
3 0
3 years ago
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Three identical lights are connected in series to a 12V battery. How does the brightness of each light compare?
kodGreya [7K]

Answer: They are identical brightness

Explanation:

If the lights are assumed to be resistance bulbs

Each light has the same current and will each drop one third of the supply voltage.  

8 0
3 years ago
An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0
3 years ago
Read 2 more answers
4. Describe how the velocity of an object changes if it undergoes uniformly acceleration motion. Can its direction change?
valentinak56 [21]

Answer:

n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

Explanation:

Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.

In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

              a_{t} = (v₂-v₁)/Δt

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

                    a_{c} = v2/R

In the general case, both the module and the address change

             a = Ra (  a_{t}^2 +   a_{c}^2)  

4 0
3 years ago
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