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3 years ago

A particle (q = +5.0 µC) is released from rest when it is 2.0 m from a charged particle which is held at

Physics

1 answer:

PilotLPTM [1.2K]3 years ago

3 0

The answer to your question is A. Have a great day

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What area must the plates of a capacitor be if they have a charge of 5.7uC and an electric field of 3.1 kV/mm between them? O 0.

kirill [66]

Answer:

Area of the plates of a capacitor, A = 0.208 m²

Explanation:

It is given that,

Charge on the parallel plate capacitor, $q = 5.7\ \mu C=5.7\times 10^{-6}\ C$

Electric field, E = 3.1 kV/mm = 3100000 V/m

The electric field of a parallel plates capacitor is given by :

$E=\dfrac{q}{A\epsilon_o}$

$A=\dfrac{q}{E\epsilon_o}$

$A=\dfrac{5.7\times 10^{-6}\ C}{3100000\ V/m\times 8.85\times 10^{-12}\ F/m}$

A = 0.208 m²

So, the area of the plates of a capacitor is 0.208 m². Hence, this is the required solution.

8 0

3 years ago

Please help with this one

olasank [31]

Answer:

OPTION A is the correct answer

6 0

2 years ago

The Department of Natural Resources has selected certain species that are on the verge of extinction due to a lack of prey. Thes

insens350 [35]

Answer:

Habitat manipulation

Explanation:

Habitat manipulation, otherwise known as ecological engineering, is a technique of promoting natural enemies within an ecosystem by making thriving conditions more suitable for them.

<em>In this case, thriving conditions for the species (which happens to be predators and hence, natural enemies) were promoted via artificial introduction of food. </em>

5 0

4 years ago

ANSWER THE 3 QUESTION WILL MARK AS BRAINLIEST THIS IS 20 POINTS :)))

PtichkaEL [24]

15) a
16) b
17) a

Hope this helps

3 0

3 years ago

Read 2 more answers

In a hydroelectric power plant, 65 m3 /s of water flows from an elevation of 90-m to a turbine-generator, where electricity is g

Mariulka [41]

The electric output of the plant is 48.19 MW

First we need to calculate the water power, it is given by the formula

WP=ρQgh

Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head

Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW

Now the overall efficiency of the hydroelectric power plant is given as

η= $\frac{electric power}{water power}$

Plugging the values in the above equation

0.84=EP/57.38

EP=48.19 MW

Therefore, the electric output of the plant is 48.19 MW.

3 0

3 years ago