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Daniel [21]
1 year ago
8

4 . A negative charge -0.450 exerts an upward 0.150N force on an unknown charge 0.250m directly below it . ( a ) what is the unk

nown charge ( magnitude and sign ) ? ( b ) what are the magnitude and direction of the force that the unknown charge exert charge exerts on the -0.450 charge ?​
Physics
1 answer:
Elodia [21]1 year ago
8 0

(a) The unknown charge kept at distance 0.250m directly below  -0.450 charge is  2.315 x 10⁻¹² C.

(b) The magnitude of the force is 29.16 x 10¹² N.

<h3>What is electrostatic force?</h3>

The electrostatic force F between two charged objects placed distance apart is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

F = kq₁q₂/r²

where k = 9 x 10⁹ N.m²/C²

Given is a negative charge -0.450 exerts an upward 0.150N force on an unknown charge 0.250m directly below it .

In equation of the magnitude of force, only the magnitude of charges are taken.

Substituting the values into the expression ,we get the charge

0.150 = (9 x 10⁹ x 0.450x q)/ (0.25)²

q = 2.315 x 10⁻¹² C

Thus, the magnitude of the unknown charge is 2.315 x 10⁻¹² C.

b)Let the unknown charge be  0.450C to exert on -450C.

Substituting the values into the expression ,we get the electrostatic force

F = kq₁q₂/r²

F = (9 x 10⁹ x 0.450x 0.450)/ (0.25)²

q = 2.315 x 10⁻¹² C

Thus, the magnitude of the force is 29.16 x 10¹² N.

Learn more about electrostatic force.

brainly.com/question/9774180

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Answer:

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Explanation:

From first Law of Thermodynamics, we know that:

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ΔU = Change in Internal Energy = -102.5 J (negative sign shows decrease in internal energy of the system)

W = Work Done in Expansion by the system = ?

Therefore,

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Now, the work done in a constant pressure condition is given by:

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W = P(Vf - Vi)

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P = Constant Pressure = (0.5 atm)(101325 Pa/1 atm) = 50662.5 Pa

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Therefore,

155 J = (50662.5 Pa)(0.058 m³ - Vi)

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<u>Vi = 0.055 m³ = 55 L</u>

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A 1.00 -kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.62a). The object has
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The distance D where the object comes to rest is 1.08.m.

<h3>What is the distance?</h3>
  • The separation of one thing from another in space; the distance or separation in space between two objects, points, lines, etc.; remoteness. The distance of seven miles cannot be accomplished in one hour of walking.
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(c) the distance D where the object comes to rest.

W_{total} =ΔKE ⇒ -0.25*1*9.8*D = 0-1/2*1*2.3^{2}

⇒D=\frac{0.5*2.3^{2} }{2.45}

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Which does not contain a lens?
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Answer:

6.77 m/s

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First, in the x direction:

Given:

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v₀ = v cos 30.8° = 0.859 v

a = 0 m/s²

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Next, in the y direction:

Given:

Δy = 0.432 m

v₀ = v sin 30.8° = 0.512 v

a = -9.81 m/s²

Δy = v₀ t + ½ at²

(0.432 m) = (0.512 v) t + ½ (-9.81 m/s²) t²

0.432 = 0.512 v t − 4.905 t²

Two equations, two variables.  Solve for t in the first equation and substitute into the second equation:

t = 3.69 / v

0.432 = 0.512 v (3.69 / v) − 4.905 (3.69 / v)²

0.432 = 1.89 − 66.8 / v²

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How much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is
mafiozo [28]

1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km

<h3>What is Speed ?</h3>

Speed is the distance travelled per time taken. It is a scalar quantity. And the S.I unit is meter per second. That is, m/s

In the given question, we want to find how much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 10^5 km.

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The parameters are;

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Speed = distance S ÷ Time t

Convert kilometer to meter by multiplying it by 1000

C = S/t

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Make t the subject of formula

t = 3.85 × 10^{8} / 3 × 10^{8}

t = 1.2833

t = 1.3 s

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