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If a point has 40 J of energy and the electric potential is 8 V, what must be the charge?

Alekssandra [29.7K]

If a point has 40 J of energy and the electric potential is 8 V, the charge must be: A. 5 C

Given the following the details;

Energy = 40 Joules

Electric potential = 8 Volts

To find the quantity of charge;

Mathematically, the quantity of charge with respect to electric potential is given by the formula;

$Quantity \; of \; charge = \frac{Energy}{Electric \; potential}$

Substituting the values into the formula, we have;

$Quantity \; of \; charge = \frac{40}{8}$

Quantity of charge = 5 Coulombs

Therefore, the quantity of charge must be 5 Coulombs.

Find more information: brainly.com/question/21808222

8 0

3 years ago

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The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th

Darya [45]

Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = $\frac{v^2}{r}$ [v = linear velocity, r = radius of circular path]

=> F = m $\frac{v^2}{r}$ ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m $\frac{v^2}{r}$ = q v B sin θ [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m $\frac{v^2}{r}$ = q v B sin 90°

=> m $\frac{v^2}{r}$ = q v B

Divide both side by v;

=> m $\frac{v}{r}$ = qB

Make v subject of the formula

v = $\frac{qBr}{m}$

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = $\frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}$

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = $\frac{1}{2}$ mv²

m = mass of proton

v = velocity of the proton as calculated above

K = $\frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )$

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

8 0

3 years ago

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo

LuckyWell [14K]

Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

a) what is the work for one cycle

b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

W = (0.3 * 8.314 * 500) * ln(5000/1000)

W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

3 0

3 years ago

What can a drop of liquid mercury be described as

jeka94

It can be described as a pure substance and an element.

7 0

3 years ago

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Which example best shows a person using the principle of force in an activity?

sleet_krkn [62]

A bowler throws a bowling ball that rolls down the lane, im like 98% sure it’s that one

6 0

2 years ago

Most people can hear sound of pitch raning from to hertz​

Most people can hear sound of pitch raning from to hertz