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2 years ago

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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 Ïƒ. 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, Ïƒ, m, and appropriate constants.

1 answer:

expeople1 [14]2 years 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 driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2

sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed $V_{0}$

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}$

Once we have the initial speed, we can isolate the final speed from following equation:

$V_{f} =V_{0} +a*t$ $V_{f}= 4.165 \frac{m}{s}$

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }$

3 0

2 years ago

Read 2 more answers

Determine the specific volume of refrigerant-134a at 1 MPa and 50°C, using (a) the ideal-gas equation of state and (b) the gener

Andrej [43]

Answer:

( a ) The specific volume by ideal gas equation = 0.02632 $\frac{m^{3} }{kg}$

% Error = 20.75 %

(b) The value of specific volume From the generalized compressibility chart = 0.0142 $\frac{m^{3} }{kg}$

% Error = - 34.85 %

Explanation:

Pressure = 1 M pa

Temperature = 50 °c = 323 K

Gas constant ( R ) for refrigerant = 81.49 $\frac{J}{kg k}$

(a). From ideal gas equation P V = m R T ---------- (1)

⇒ $\frac{V}{m}$ = $\frac{R T}{P}$

⇒ Here $\frac{V}{m}$ = Specific volume = v

⇒ v = $\frac{R T}{P}$

Put all the values in the above formula we get

⇒ v = $\frac{323}{10^{6} }$ ×81.49

⇒ v = 0.02632 $\frac{m^{3} }{kg}$

This is the specific volume by ideal gas equation.

Actual value = 0.021796 $\frac{m^{3} }{kg}$

Error = 0.02632 - 0.021796 = 0.004524 $\frac{m^{3} }{kg}$

% Error = $\frac{0.004524}{0.021796}$ × 100

% Error = 20.75 %

(b). From the generalized compressibility chart the value of specific volume

$\frac{V}{m}$ = v = 0.0142 $\frac{m^{3} }{kg}$

The actual value = 0.021796 $\frac{m^{3} }{kg}$

Error = 0.0142 - 0.021796 = $\frac{m^{3} }{kg}$

% Error = $\frac{- 0.0076}{0.021796}$ × 100

% Error = - 34.85 %

3 0

2 years ago

. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it

kap26 [50]

Answer:

29.4 N/m

0.1

Explanation:

a) From the restoring Force we know that :

F_r = —k*x

the gravitational force :

F_g=mg

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg </em>

<em>x1 = -0.75 m </em>

<em>x2 = -0.2 m </em>

<em>g = 9.8 m/s^2 </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m

Plugging known infromation to get :

l= x1 — (F_1/k)

= 0.1

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2 years ago

Serena is a research student who has conducted an experiment on the discoloration of marble. Read about Serena’s experiment. The

sergij07 [2.7K]

The two flaws in her experiment’s design are

<span>- She introduced at least one confounding variable.</span> <span>- She tried to test multiple hypotheses at a time</span>

In the above mentioned experiment she had to have four samples to prove four hypotheses, each one separately and not to mix two hypotheses in an alone sample, that what it brings as consequence is the confusion.

3 0

2 years ago

A lamp uses a 230 V mains supply and transfers 96 J of energy every second. Work out the current through the lamp. Give your ans

sertanlavr [38]

Answer:

0.4 A

Explanation:

From the question,

Electric power = Voltage×current

P = VI.......................... Equation 1

Make I the subject of the equation

I = P/V..................... Equation 2

Given: P = 96 J/s, V = 230 V.

Substitute into equation 2

I = 96/230

I = 0.4 A.

Hence the current is 0.4 A

8 0

2 years ago