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ahrayia [7]
4 years ago
4

All charged objects create an electric field around them. What two factors determine the strength of two electric fields upon th

e charged objects creating them?
A.) how charge the object creating the field is and the mass of the two charged objects


B.) how charged object creating the field is and the distance between the two charged objects

C.) how charged object creating the field is and whether the objects are positively or negatively charged

D.) how charged object creating the field is in the density of the two charged objects
Physics
1 answer:
Ierofanga [76]4 years ago
8 0
The answer is C.<span> how charged object creating the field is and whether the objects are positively or negatively charged.
</span>
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A cube made out of wood has a mass of 1.2 kg and the density of the wood is 400kg/m3.What is the length of the side of the cube?
LenaWriter [7]

Answer:

≈ 0.144 m

Explanation:

Density is mass divided by volume:

D = M / V

Solving for volume:

V = M / D

Given M = 1.2 kg and D = 400 kg/m³:

V = 1.2 kg / (400 kg/m³)

V = 0.003 m³

Volume of a cube is the side length cubed:

V = s³

Therefore:

s³ = 0.003 m³

s ≈ 0.144 m

Round as needed.

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3 years ago
Explain the energy transformations that occur when breaking in a hybrid<br> vehicle.
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it occur kinetic energy

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A fire truck has a searchlight with a resistance of 60 (ohm) which is placed across a 24-V battery. What is the current in this
Mariulka [41]

Answer: 0.4 Amps

Explanation:

Voltage of battery = 24 Volts

Current I = ?

Resistance of searchlight (R)= 60ohms (Ω is the symbol for ohms)

Then, apply the formula for ohms law

Voltage = Current x resistance

i.e V = I x R

24V = I x 60Ω

I = 24V / 60Ω

I = 0.4 Amps (Amps is the unit of current)

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Deep currents circulate seawater around the globe due to differences in the density of water at different locations. Two factors
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How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 20.0 cm in diameter to p
Rufina [12.5K]

Answer:

1.007 × 10^(10) electron

Explanation:

We are given;

Electric Field;E = 1450 N/C

Diameter;d = 20 cm = 0.2 m

So, Radius: r = 0.2/2 = 0.1 m

Formula for Electric field just outside the sphere is given by the formula;

E = kq/r²

Where;

E is the magnitude of the electric field. q is the magnitude of the point charge r is distance from the point charge

k is a constant with a value of 9 x 10^(9) N.m²/C²

Making q the subject, we have;

q = Er²/k

Thus,

q = 1450 × 0.1²/(9 × 10^(9))

q = 1.61 × 10^(-9) C

Now, total charge q is also given by the formula;

q = Ne

Where;

e is charge on electron which is 1.6 × 10^(-19)

N is number of excess electrons

Making N the formula, we have;

N = q/e

N = (1.61 × 10^(-9))/(1.6 × 10^(-19))

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