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2 years ago

What is the magnitude of a point charge in coulombs whose electric field 51 cm away has the magnitude 2.5 n/c?

Physics

1 answer:

Tema [17]2 years ago

7 0

The magnitude of a point charge is 0.189 x 10⁻⁹C.

<h3>Steps</h3>

We have stated that the point charge's electric field is E=2.5 N/C.

Distance from point charge R = 51 cm = 0.51 m

We are aware that the electric field resulting from a point charge is given as E = 1 / 4π∈₀ × Q / R²

so 2.5 = 9 x 10⁹ Q/ 0.68²

The magnitude of a point charge is 0.189 x 10⁻⁹C.

The electric force per unit charge is referred to as the electric field. It is assumed that the field's direction corresponds to the force it would apply to a positive test charge.

From a positive point charge, the electric field radiates outward, and from a negative point charge, it radiates in.

learn more about electric field here

brainly.com/question/14372859

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The first moment of the total cross-sectional area taken about the neutral axis must be?

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The first moment of the total cross sectional area taken about the neutral axis must be zero.

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2 years ago

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What determines how severe an electric shock will be?

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3 years ago

A high diver of mass 74.0 kg jumps off a board 9.00 m above the water. If his downward motion is stopped 2.50 seconds after he e

stiv31 [10]

Answer:

1120 N

Explanation:

The velocity with which he hits the water can be found with kinematics:

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-9.00 m)

v = -13.3 m/s

Or it can be found with conservation of energy.

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × -9.8 m/s² × -9.00 m)

v = -13.3 m/s

Sum of forces on the diver after he hits the water:

∑F = ma

F − mg = m Δv/Δt

F − (74.0 kg) (9.8 m/s²) = (74.0 kg) (0 m/s − (-13.3 m/s)) / (2.50 s)

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3 years ago

In positron-emission tomography (PET) used in medical research and diagnosis, compounds containing unstable nuclei that emit pos

arlik [135]

Answer:

Energy of gamma ray = 506250 eV

Explanation:

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This means that their energies will be the same.

Thus; E_e = E_p

Now, since electron and positron annihilate to form gamma(γ) particle, then using work energy principle, we have;

E_γ + E_γ = E_e + E_p

Thus;

2E_γ = E_e + E_p

Since E_e = E_p, we now have;

2E_γ = 2E_e

Thus;

E_γ = E_e

Since all the energy of the electron is converted, then from Einstein's relativity theory, this implies that;

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And m is given as 9 × 10^(-31) kg

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Since E_γ = E_e

Thus;

E_γ = 810 × 10^(-16) J

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(810 × 10^(-16) × 1)/(1.6 × 10^(-19)) = 506250 eV

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3 years ago