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Strike441 [17]
1 year ago
5

A small sphere with mass m carries a positive charge q and is attached to one end of a silk fiber of length L. The other end of

the fiber is attached to a large vertical insulating sheet that has a positive surface charge density σ. Assume that the sphere is in equilibrium and find the angle that fiber makes with the vertical sheet. Express your answer in terms of the variables q, σ, m, and appropriate constants.
Physics
1 answer:
expeople1 [14]1 year ago
7 0

Here in the given situation there is a constant electric field due to large sheet

This constant electric field is given as

E = \frac{\sigma}{2\epsilon_0}

now we know that force on the ball due to this electric field will be in horizontal direction given as

F = qE

F = \frac{\sigma q}{2\epsilon_0}

now this horizontal force will be balanced by horizontal component of the tension in the string

Tsin\theta = \frac{\sigma q}{2 \epsilon_0}

now similarly we can say that vertical component of tension force in the string will balance the weight of small sphere

Tcos\theta = mg

now from above two equations we will have

tan\theta = \frac{\sigma q}{2\epsilon_0 mg}

\theta = tan^{-1}\frac{\sigma q}{2\epsilon_0 mg}

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The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You
tatiyna

Answer:

Explanation:

Given

B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)

Em wave is in the form of

B=B_0\sin (kx-\omega t)

where \omega =frequency\ of\ oscillation

k=wave\ constant

B_0=Maximum\ value\ of\ Magnetic\ Field

Wave constant for EM wave k is

k=1.05\times 10^7 m^{-1}

Wavelength of wave \lambda =\frac{2\pi }{k}

\lambda =\frac{2\pi }{1.05\times 10^7}

\lambda =5.98\times 10^{-7} m

7 0
2 years ago
A technician is troubleshooting a problem. The technician tests the theory and determines the theory is confirmed. Which of the
vladimir1956 [14]

Answer:

b)  Document lessons learned.

Explanation:

First he should do documentation

then C

then D

then A

3 0
2 years ago
A professor determined the relationship between the time spent studying (in hours) and performance on an exam.
lubasha [3.4K]
Lets write again formula for determening Ann's performance.

P = 70.443 + 4,885*t

where t is in hours. This is equation with P(t) which means that P only depends on variable t. If we express t=2.6 in formula we will find her expected performance.

P = 83.144

Now, since it says that she scored 16 points less than expected we need to find value of P-16

P - 16 = 67.144

After round we get that the answer is 67
3 0
2 years ago
A barbellbell is loaded with two 20 kg plates on its right side and two 20kg plates on its left side. The barbell is 2.2m long,
Aleks [24]

Answer:

<h2>Center of gravity of the system of all masses is 105 cm from the left end of the rod</h2>

Explanation:

Let the position of Left end of the rod is our reference

So here we will have

m_1 = 20 kg

x_1 = 20 cm

m_2 = 20 kg

x_2 = 30 cm

m_3 = 20 kg

x_3 = 110 cm

m_4 = 20 kg

x_4 = (220 - 40) = 180 cm

m_5 = 20 kg

x_5 = (220 - 35) = 185 cm

so we will have

x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4 + m_5x_5}{m_1 + m_2 + m_3 + m_4 + m_5}

x_{cm} = \frac{20(20 + 30 + 110 + 180 + 185)}{20 + 20 + 20 + 20 + 20}

x_{cm} = 105 cm

7 0
2 years ago
A car speeds up from 13 m/s to 23 m/s in 30 seconds. What is the acceleration of the car?
ra1l [238]
Well, <span>v = u + a×t is the equation.</span>
<span>
v: final velocity, which is 23 m/s in this equation.</span>
<span>u: initialo velocity = 13 m/s </span>
<span>a: acceleration = ? </span>
<span>t: time = 30s 
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Your equation would be...

<span>23 = 13 + a×30 </span>
<span>a = (23 - 13) / 30 </span>
<span>a = 1 / 3 </span>
<span>a = 0.333 m/s</span>
7 0
2 years ago
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