PHYSICS HIGH SCHOOL

Name three devices that can store electric potential energy

Answers

Answer 1
Answer:

Potential energy, unlike kinetic, is stored energy. This energy comes from different sources. An object with potential energy also has the ability to produce energy if something triggers it. Example, a charged battery, a car on top of a hill, and capacitors.

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A 755 N diver drops from a board 10.0 m above the water’s surface. Find the mechanical energy of the system, and find the diver’s speed 5.00 m above the water’s surface. Then find the diver’s speed just before striking the water.

Answers

This is a conservation of energy problem. E0=E1

Energy initial = mgh
Energy final = KE = 1/2mv^2

A charge of 8.5 × 10–6 C is in an electric field that has a strength of 3.2 × 105 N/C. What is the electric force acting on the charge?

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ok
here is your anwer
O hope it is useful for you

Answer:

2.7N

Explanation:

The metric unit of force is the

Answers

Newtons. Force is mass times acceleration. Mass is measured in kilograms (kg) and acceleration is measured in meters per second squared (m/s^2.) These units (kg and m/s^2) multiplied together like in the equation force equals mass times acceleration (F=ma) gives a product with Newtons as the unit. 

Is it ok for some help pwease?What is the area of calm winds near the equator called?


doldrums


polar easterlies


trade winds


horse latitudes

Answers

Answer:

doldrums

Explanation:

Just need to type something more to be able to post

A planet moves fastest when it is ___________________ to the Sun and slowest when it is ___________________ from the Sun. Choose one: A. farthest; closest B. at the nearest focus; at the farthest focus C. closest; farthest

Answers

Answer:

option (c) is correct

Explanation:

A planet moves fastest when it is closest to the Sun and slowest when it is farthest from the Sun.

Answer:

fastest when closest to the Sun, slowest when it is farthest

Explanation:

Plato/Edmentum

The parallel plates in a capacitor, with a plate area of 8.50 cm2 and an air-filled separation of 3.00 mm, are charged by a 6.00 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 8.00 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates

Answers

Answer:

(a). The charge is 1.5045*10^(-11)\ C

(b). The initial stored energy is 4.5135*10^(-11)\ J

(c).  The final stored energy is 12.036*10^(-11)\ J

(d). The work required to separate the plates is 7.5225*10^(-11)\ J

Explanation:

Given that,

Area = 8.50 cm²

Distance = 3.00 mm

Potential = 6.00 V

Distance without discharge = 8.00 mm

(a). We need to calculate the capacitance

Using formula of capacitance

C_(1)=(\epsilon_(0)A)/(d)

Put the value into the formula

C_(1)=(8.85*10^(-12)*8.5*10^(-4))/(3.00*10^(-3))

C_(1)=2.5075*10^(-12)\ F

We need to calculate the charge

Using formula of charge

Q=CV

Put the value into the formula

Q=2.5075*10^(-12)*6.00

Q=1.5045*10^(-11)\ C

(b). We need to calculate the initial stored energy

Using formula of initial energy

E_(i)=(1)/(2)* CV^2

E_(i)=(1)/(2)*2.5075*10^(-12)*36

E_(i)=4.5135*10^(-11)\ J

(c). We need to calculate the capacitance

Using formula of capacitance

C_(2)=(\epsilon_(0)A)/(d)

Put the value into the formula

C_(2)=(8.85*10^(-12)*8.5*10^(-4))/(2*8.00*10^(-3))

C_(2)=9.403*10^(-13)\ F

We need to calculate the final stored energy

Using formula of initial energy

E_(f)=(1)/(2)* (Q^2)/(C)

E_(f)=(1)/(2)*((1.5045*10^(-11))^2)/(9.403*10^(-13))

E_(f)=12.036*10^(-11)\ J

(d). We need to calculate the work done

Using formula of work done

W=E_(f)-E_(i)

Put the value in the formula

W=12.036*10^(-11)-4.5135*10^(-11)

W=7.5225*10^(-11)\ J

Hence, (a). The charge is 1.5045*10^(-11)\ C

(b). The initial stored energy is 4.5135*10^(-11)\ J

(c).  The final stored energy is 12.036*10^(-11)\ J

(d). The work required to separate the plates is 7.5225*10^(-11)\ J
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