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den301095 [7]
3 years ago
14

A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.442 and t

he coefficient of kinetic friction is 0.240. How much force is needed to begin moving the book?
How much force is needed to keep the book moving at constant speed once it begins moving?
Physics
1 answer:
photoshop1234 [79]3 years ago
8 0

Answer:

a) 8.80 N

b) 4.78 N

Explanation:

The weight of the book is

W = mg = 2.03\times9.81 = 19.9143 \text{ N}

By Newton's third law, the reaction from the table, R, is equal to its weight.

R = 19.9143 N

From the law of solid friction,

F = \mu R

F is the maximum frictional force and \mu is the coefficient of friction.

a) The force needed to begin moving the book is that needed to overcome static friction

F = 0.442\times19.9143 = 8.80\text{ N}

b) The force needed to move the book at constant speed once it begins moving is that needed to overcome kinetic friction

F = 0.240\times19.9143 = 4.78\text{ N}

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Besides, we have to consider the resitance equivalent for a parallel connection. This is given by:

1/Re=1/R1+1/R2

If we connect the same resistance, the equivalent resistance is R/2.

Initlally  the current is 1.5 A when one resistance is connected to the batttery. When a second resistance with the same value is connected in parallel to the battery, we have V=Re*Ifinal= (R/2)*Ifinal

also we know that V=R*Iinitial so Iinitial=V/R

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4 years ago
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Answer:

0.0025 sec

Explanation:

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2 years ago
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You throw a ball straight up. The ball has an initial speed of 11.2 m/s when it leaves your hand What is the maximum height the
Jet001 [13]

Answer:

Explanation:

Given

initial velocity u=11.2\ m/s

At maximum height velocity of ball is zero

using

v^2-u^2=2as

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a=acceleration

s=displacement

(0)^2 -(11.2)^2=2\times (-9.8)\times (s)

s=\frac{11.2^2}{2\times 9.8}

s=6.4

time taken by the ball to reach the maximum height

v=u+at

0=11.2-9.8\times t

t=\frac{11.2}{9.8}

t=1.142\ s

At t=2\ s height of ball is

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h=11.2\times 2-\frac{1}{2}9.8\times (2)^2

h=22.4-19.6

h=2.8\ m

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3 years ago
a constant current of 3 a for 4 hours is required to charge an automotive battery.if the terminal voltage is 10 t/2 v, where t i
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Current (i) = 3A

Time interval (t) = 4 hrs = 14400secs

As we know ,

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Mathematically , it can be written as

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Q = i × t

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By dividing the value of the current passed through the circuit by the time for which the charge is passed gives out the value of the charge passed .

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Answer:

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