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1 year ago

12

A 22.8 kg rocking chair begins to slide across the carpet when the push reaches 57.0 N. What is the coefficient of static fricti

on?

1 answer:

vfiekz [6]1 year ago

7 0

Answer:

0.255

Explanation:

The following data were obtained from the question:

Force (F) = 57 N

Mass (m) = 22.8 Kg

Coefficient of static friction (µ) =...?

Next, we shall determine the normal reaction (R). This is illustrated below:

Mass (m) = 22.8 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal reaction (R) =?

R = mg

R = 22.8 x 9.8

R = 223.44 N

Finally, we can obtain the coefficient of static friction (µ) as follow:

Force (F) = 57 N

Normal reaction (R) = 223.44 N

Coefficient of static friction (µ) =...?

F = µR

57 = µ x 223.44

Divide both side by 223.44

µ = 57/223.44

µ = 0.255

Therefore, the coefficient of static friction (µ) is 0.255.

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What voltage is required to move 6A through 20?<br><br>

Marina CMI [18]

Answer:

120V

Explanation:

Given parameters:

Current = 6A

Resistance = 20Ω

Unknown:

Voltage = ?

Solution:

According to ohms law;

V = IR

Where V is the voltage

I is the current

R is the resistance

Now, insert the parameters and solve;

V = 6 x 20 = 120V

7 0

1 year ago

Write the equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and

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Answer:

The newton’s second law is F=ma

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Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

Velocity :

The velocity is equal to the rate of position of the object.

$v=\dfrac{dx}{dt}$ ....(I)

Acceleration :

The acceleration is equal to the rate of velocity of the object.

$a=\dfrac{dv}{dt}$ ....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

$F=\dfrac{d(p)}{dt}$

Put the value of p

$F=\dfrac{d(mv)}{dt}$

$F=m\dfrac{dv}{dt}$

Put the value from equation (II)

$F=ma$

This is newton’s second laws.

Gravitational force :

The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

$F=\dfrac{Gm_{1}m_{2}}{r^2}$

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

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2 years ago

An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its

FromTheMoon [43]

Answer:

The terminal speed of this object is 12.6 m/s

Explanation:

It is given that,

Mass of the object, m = 80 kg

The magnitude of drag force is,

$F_{drag}=12v+4v^2$

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.

$F_{drag}=mg$

$12v+4v^2=80\times 9.8$

$4v^2+12v=784$

On solving the above quadratic equation, we get two values of v as :

v = 12.58 m/s

v = -15.58 m/s (not possible)

So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.

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1 year ago

Wrapping paper is being unwrapped from a 5.0-cm radius tube, free to rotate on its axis. if it is pulled at the constant rate of

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So the equation for angular velocity is

Omega = 2(3.14)/T

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In order to find T, the tangential velocity is

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2 years ago

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

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$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

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$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

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2 years ago