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Alex777 [14]
3 years ago
10

Two 2.0-cm-diameter insulating spheres have a 6.60 cm space between them. One sphere is charged to + 76.0 nC , the other to - 30

.0 nC . Part A What is the electric field strength at the midpoint between the two spheres? Express your answer with the appropriate units.
Physics
1 answer:
e-lub [12.9K]3 years ago
8 0

Answer:

5.2\times 10^5N/C

Explanation:

Since the two charged bodies are symmetric, we can calculate the electric field taking both of them as point charges.

This can be easily seen if we use Gauss's law, \int{E} \, dA=\frac{Q_{enclosed}}{\epsilon_o}

We take a larger sphere of radius, say r, as the Gaussian surface. Then the electric field due to the charged sphere at a distance r from it's center is given by,

E=\frac{1}{4\pi r^2} \frac{Q_{enclosed}}{\epsilon_o}

which is the same as that of a point charge.

In our problem the charges being of opposite signs, the electric field will add up. Therefore,

E_{total}=\frac{1}{4\pi\epsilon_o}\frac{q_1+q_2}{r^2}= (9\times10^9) \frac{(76+30)\times10^{-9}}{((1+3.3)\times10^{-2})^2}N/C =5.2\times10^5N/C

where, r = distance between the center of one sphere to the midpoint (between the 2 spheres)

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ivolga24 [154]

Explanation:

Given data:

d = 30 mm = 0.03 m

L = 1m

S_{y} = 70 Mpa

Δd = -0.0001d

Axial force = ?

validity of elastic deformation assumption.

Solution:

O'₂ = Δd/d = (-0.0001d)/d = -0.0001

For copper,

v = 0.326      E = 119×10³ Mpa

O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶

∵δ = F.L/E.A    and σ = F/A so,

σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa

F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN

S_{y} = 70 MPa > σ = 36.5 MPa

∵ elastic deformation assumption is valid.

so the answer is

F = 25800 K N            and     S_{y} > σ

3 0
3 years ago
A bullet B of mass mB traveling with a speed v0 = 1400 m/s ricochets off a fixed steel plate A of mass mA. Let mA ≫ mB so that i
Greeley [361]

Answer:

Rebounce angle is 345°

Rebounce speed is 989.95m/s

Explanation:

Calculate the x  component of the velocity of the bullet before impact by using the following relation:

Vbx= Vb Cos thetha

Here,  is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°

Substituting

Vbx = Cos15 ×1400 = 1352.30m/s

Calculate the y component using the relation:

Vby = Vo Sin theta

Vby = sin 15° × 1400

Vby = 362.35m/s

The rebounce angle = 360 - incidence angle

Rebounce angle =( 360 - 15)° = 345°

The rebound speed V' = Vby - Vbx

V' = (1352.30 - 362.35)m/s

V' = 989.95 m/s

5 0
3 years ago
If a sound waves frequency is 100Hz. what is its period
agasfer [191]

Period = (1/frequency) .

If frequency is 100 per second, then

     Period = (1) / (100 per second)  =  0.01 second .

4 0
3 years ago
Determine the amount of time for polonium-210 to decay to one fourth its original quantity. The half-life of polonium-210 is 138
Tju [1.3M]

Answer:

276 days

Explanation:

1/4 th of the original means <u><em>2 half lives</em></u>

1 half life = 138 days

So,

2 half lives = 276 days

5 0
3 years ago
when the piston of a fountain pen with a nib is dipped into ink and and the air is released by pressing it, the ink fills in the
____ [38]

Answer: Please find the answer in the explanation

Explanation: According to the Newton 3rd law of motion which state that;

In every action, there will be equal and opposite reaction.

when the piston of a fountain pen with a nib is dipped into ink and and the air is released by pressing it, a force is applied which act on the molecules of the ink by pushing the molecules. When the force is released, there will be a reaction force by pulling the molecules of the ink into the pen. Thereby the ink fills in the pen. 

7 0
3 years ago
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