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Stells
2 months ago
7

What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can a

nd the ground is 0.76.

Physics
1 answer:
serg [3.5K]2 months ago
8 0

Response:

T = 38.38 N

Explanation:

In this case

mass of can = m = 3 kg

g= 9.8 m/sec²

angle θ = 40°

As illustrated, we observe the vertical and horizontal components of the tension force T.

If the can is on the verge of slipping, then the horizontal component of the tension force should equal the frictional force.

First, we calculate the frictional force

Fs = μ R

where

μ = 0.76

R = weight of the can = mg = 3 × 9.8 = 29.4 N

Now, regarding the horizontal component of tension, we have

Tx = T cos 40° = T × 0.7660 N

which leads to T × 0.7660 = 29.4

Thus, T = 38.38 N

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A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "
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Answer:

a) Q = C*emf

b)  Decrease in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

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a) The resistors and capacitors are linked in series with the battery

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V_{c}and

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.........................(1)

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Integrating V_{c}and V_{R}into equation (1)

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I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\

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I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}

d) Note: The initial charge on the capacitor equals C * emf

Following the insertion of the plastic, the new charge will be:

Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf

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