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Vikki [24]
3 years ago
14

Two tiny beads are 25 cm apart with no other charged objects or fields present. Bead A has a net charge of magnitude 10 nC and b

ead B has a net charge of magnitude 1 nC. Which one of the following statements is true about the magnitudes of the electric forces on these beads?A. The force on A is 100 times the force on B.B. The force on B is 10 times the force on A.C. The force on B is 100 times the force on A.D. The force on A is 10 times the force on B.E. The force on A is exactly equal to the force on B.
Physics
1 answer:
Alona [7]3 years ago
6 0

Answer:

E. The force on A is exactly equal to the force on B.

Explanation:

The force between two charges is given by

F=\dfrac{kq_1q_2}{r^2}

where

q_1 = Charge on particle 1

q_2 = Charge on particle 2

r = Distance between the charges

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

F=\dfrac{8.99\times 10^9\times 10\times 10^{-9}\times 1\times 10^{-9}}{(25\times 10^{-2})^2}\\\Rightarrow F=0.0000014384\ N

This force will be exerted on both the charges equally.

So, The force on A is exactly equal to the force on B

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The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
Nikolay [14]

Answer:

Index of expansion: 4.93

Δu = -340.8 kJ/kg

q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

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p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

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p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

The variation of specific internal energy is:

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Δu = 0.71 * (20 - 500) = -340.8 kJ/kg

The first law of thermodynamics

q = l + Δu

The work will be the expansion work

l = p2*v2 - p1*v1

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7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

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a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres​
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The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

       \text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}

Similarly,

      \text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}

We know the below formulas,

          \text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}

          \text {volume of wire}=\pi \times r^{2} \times l

When equating both the equations, we can find length of wire as below, where \pi=\frac{22}{7}

          \frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l

         \frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l

The \pi value gets cancelled as common on both sides, we get

           \frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l

The 10^{-6} value gets cancelled as common on both sides, we get

           l=4 \times 9=36 m

7 0
3 years ago
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