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dlinn [17]
3 years ago
9

When you push a 2.00 kg book resting on a tabletop it takes 4.60 N to start the book sliding. What is the coefficient of static

friction between the book and the tabletop? g
Physics
1 answer:
natali 33 [55]3 years ago
3 0

The coefficient of static friction is 0.234.

Answer:

Explanation:

Frictional force is equal to the product of coefficient of friction and normal force acting on any object.

So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.

Normal force = mass * acceleration due to gravity

Normal force = 2 * 9.8 = 19.6 N.

And the frictional force is given as 4.6 N, then

Coefficient of static friction = Frictional force/Normal force

Coefficient of static friction = 4.6 N / 19.6 N = 0.234

So the coefficient of static friction is 0.234.

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Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?
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The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

h = \frac{v^{2}}{2g}

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

\frac{mv^{2}}{2} = mgh

h = \frac{v^{2}}{2g}

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

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Problem: The frequency of an FM radio station is 89.3 MHz. Calculate its period. Part B: From the Library, select the general eq
vekshin1

Answer:

Time period, T=1.11\times 10^{-8}\ s

Explanation:

We have,

The frequency of an FM radio station is 89.3 MHz.

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The reciprocal of frequency is called time period of a wave. It can be given by :

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3 years ago
a spiral spring of natural length 10.0cm has a scale pan hanging freely in its tower ends . When an object of mass of 20g is pla
weqwewe [10]

Answer:

Explanation:

There seems to be a typo in the problem statement.  It says the spring stretches to a shorter length after more mass is added.  Please check the problem statement.  I'm going to do the calculations assuming that the first length should be 11.80 cm and the second length should be 12.05 cm.

Hooke's law states that the force needed to compress or extend a linear spring is:

F = kΔx, where k is the stiffness and Δx is the displacement.

When a 20g object is placed in the pan, the spring stretches to a length of 11.80 cm.  The force of the spring is counteracting the weight of both the pan and the object. Therefore:

(m + 0.020) g = k (0.1180 - 0.100)

And when another 30g object is placed in the pan, the spring stretches to a length of 12.05 cm.

(m + 0.020 + 0.030) g = k (0.1205 - 0.100)

We now have two equations and two variables.  If we divide the second equation by the first equation:

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0.0180 (m + 0.050) = 0.02050 (m + 0.020)

0.0180 m + 0.0009 = 0.02050 m + 0.00041

0.00049 = 0.0025 m

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The pan has a mass of 0.196 kg, or 196 g.

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<span>Unsafe passes are passes with restricted line of sight, passes with cross traffic, narrow passes which are unsafe. 
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