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Bezzdna [24]
3 years ago
7

11 e) Can a conductor be given limitless charge? Obtain the equivalent resistance of several resistors if (a) they are in series

, (b) they are in parallel​
Physics
1 answer:
horsena [70]3 years ago
6 0

Answer:

(e) no

(a) Rs = R' + R'' + R'''

(b) 1/Rp = 1/R' + 1/R'' + 1/R'''

Explanation:

11 e)

Practically it is not possible to give limitless charge to a conductor. It depends to the number of valence electrons.

(a) When the three resistances R'. R'' and R''' is in series combination.

Let they are connected to the voltage V and the current in each resistance is I.

According to Ohm's law

Voltage across R', V' = I R'

Voltage across R'', V'' = I R''

Voltage across R''', V''' = I R'''

So, let the equivalent resistance is Rs.

I Rs = I R' + I R'' + I R'''

Rs = R' + R'' + R'''

(b)

When the three resistances R'. R'' and R''' is in parallel combination.

Let they are connected to the voltage V and the current in each resistance is I', I''. I'''.

Current in R', I' = V/R'

Current in R'', I'' = V/R''

Current in R''', I''' = V/R'''

The equivalent resistance is Rp.

V/Rp = V/R' + V/ R'' + V/R'''

1/Rp = 1/R' + 1/R'' + 1/R'''

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A ball starts from rest and accelerates at 0.460 m/s2 while moving down an inclined plane 8.85 m long. When it reaches the botto
Ivahew [28]

Answer:

V=2.8534m/s \approx 3m/s

Explanation:

From the question we are told that

Acceleration a=0.460m/s

Distance traveled S_1=8.5m

Distance traveled S_2=15.6m

Generally the velocity of the ball at the bottom of the first plane V is mathematically given by

v^2=2as

v=\sqrt{2as}

V=\sqrt{2*0.460*8.85}

V=2.8534m/s \approx 3m/s

6 0
3 years ago
Two charges, - Q0 and - 4Q0 are a distance d apart. These two charges are free to move but do not because there is a third charg
TEA [102]

To solve this problem we will use the equilibrium conditions in the Electrostatic Forces. In turn, we will use the concept formulated from Coulomb's laws to determine the intensity of the Forces and make the respective considerations.

Our values for the two charges are:

q_1 = -Q_0

q_2 = -4Q_0

As a general consideration we will start by determining that they are at a unit distance (1) separated from each other. And considering that both are negative charges, they will be subjected to repulsive force. Said equilibrium compensation will be achieved only by placing a third force between the two.

Let the third charge be q_3 = +Q is placed at a distance x from q_1

F_{1,3} = \frac{k(-Q_0)(+Q)}{x^2}

The force on q_3 due to q_2 is

F_{2,3} = \frac{k(-4Q_0)(+Q)}{1-x^2}

The condition of equilibrium is

F_{1,3} = F_{2,3}

\frac{k(-Q_0)(+Q)}{x^2}= \frac{k(-4Q_0)(+Q)}{1-x^2}

\frac{1}{x^2} = \frac{4}{(1-x)^2}

x = 0.331 from q_1

To find the magnitude of q_3 we use F_{1,2} = F_{1,3}

\frac{k(-Q_0)(4Q_0)}{1^2}= \frac{k(-Q_0)(Q)}{0.331^2}

Q = 0.43Q_0

The magnitude of the third charge must be 0.43 the first charge Q_0

3 0
4 years ago
A convex lens of focal length 25 cm is in contact with a concave lens of focal length 50 cm. The equivalent focal length of this
Gwar [14]

The equivalent focal length of this combination-lenses should be +50 cm.

Given:

Focal length of convex lens, f₁ = +25 cm

Focal length of concave lens, f₂ = -50 cm

Calculation:

We know that the power of a lens is given as:

P = 1 / f

where, f is the focal length of lens

Now, the power of convex lens can be calculated as:

P₁ = 1/f₁

  = 1/(25×10⁻² m)

  = 100/25

  = +4 D

Similarly, the power of concave lens can be calculated as:

P₂ = 1/f₂

    = (1/-50×10⁻² m)

    = -100/50

    = -2 D

Thus, the power of the combined lenses can be calculated as:

P₍₁₊₂₎ = P₁ + P₂

       = +4 D - 2 D

       = +2 D

Now, the combined focal length of the lens can be calculated as:

f₍₁₊₂₎ = 1 / P₍₁₊₂₎

      = 100 / 2 D

      = +50 cm

Therefore, the equivalent focal length of this combination-lenses will be +50 cm.

Learn more about mirrors and lenses here:

<u>brainly.com/question/3209252</u>

#SPJ4

6 0
2 years ago
The hypotenuse of a right triangle is 29.0 centimeters. The length of one of its legs is 20.0 centimeters. What is the length of
Kisachek [45]

square root (29 squared - 20 squared)

Pythagoras' theorem

5 0
3 years ago
Read 2 more answers
Two cars are traveling around identical circular racetracks. Car A travels at a constant speed of 20 m/s. Car B starts at rest a
Tomtit [17]

Answer:

b. it has the same centripetal acceleration as car A.

Explanation:

According to the question, the data provided is as follows

Constant speed of car A = 20 m/s

Constant tangential acceleration until its speed is 40 m/s

Based on the above information, the true statement is the same centripetal acceleration as car A because

As we know that

Centripetal acceleration is

= \frac{V^2}{r}

where,

V^2 = velocity

r = radius of the path

Now if both car A and car B moving in the same or identical circular path having the same velocity so in this case there is the same centripetal acceleration for that particular time

hence, the second option is correct

3 0
4 years ago
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