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Stolb23 [73]
3 years ago
10

A book of mass 7 kg rests on a plank. You tilt one end of the plank and slowly increase the angle of the tilt. The coefficient o

f static friction between the book and the plank is 0.32. What is the maximum angle of tilt for which the book will remain stationary and not slide down the plank
Physics
2 answers:
VARVARA [1.3K]3 years ago
5 0

Answer:

17.74°

Explanation:

For an inclined surface,

F = μmgcosΦ................ Equation 1

F' = mgsinΦ................. Equation 2

Where F = friction force, F' = down slope force, m = mass of the book, g = acceleration due to gravity, μ = coefficient of static friction, Φ = maximum angle of tilt.

For the book to remain stationary and not slide.

F = F'

μmgcosΦ = mgsinΦ

μ = cosΦ/sinΦ

μ = tanФ

make Ф the subject of the equation

Ф = tan⁻¹(μ).................... Equation 3

Given: μ = 0.32

Substitute into equation 3

Ф = tan⁻¹(0.32)

Ф = 17.74°

MAXImum [283]3 years ago
5 0

To solve this problem we will apply the concepts related to the Friction force and the force induced by gravity. Since the displacement is in angular mode, the component of the horizontal force of friction will be equivalent to the component of the vertical force of gravity. For balance to exist, both must be equal

F_{fx} = F_{gy}

Where,

F_{fx} = \mu mg cos\theta

F_{gy} = mg sin \theta

m = Mass

g = Gravitational acceleration

\mu = Constant of friction

Then,

m g sin \theta = \mu m g cos \theta

sin \theta = mu cos \theta

tan \theta = mu

tan\theta = 0.32

\theta = 17.74\°

Therefore the maximum angle  of tilt for which the book will remain stationary and not slide down the plank is 17.74°

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Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree
kvv77 [185]

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is    

                    I_f_3 = 27.57 W/cm^2

2 When third  polarizer is removed the intensity after it passes through the stack is    

                I_f_2 = 102.24 W/cm^2

Explanation:

  From the question we are told that

       The angle of the second polarizing to the first is  \theta_2 = 21^o  

        The angle of the third  polarizing to the first is     \theta_3 = 61^o

        The unpolarized light after it pass through the polarizing stack   I_u = 60 W/cm^2

Let the initial intensity of the beam of light before polarization be I_p

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

                     I_1 = \frac{I_p}{2}

Now according to Malus’ law the  intensity of light that would emerge from the second polarizing filter is mathematically represented as

                    I_2 = I_1 cos^2 \theta_1

                       = \frac{I_p}{2} cos ^2 \theta_1

The intensity of light that will emerge from the third filter is mathematically represented as

                  I_3 = I_2 cos^2(\theta_2 - \theta_1 )

                          I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]

making I_p the subject of the formula

                  I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}

    Note that I_u = I_3 as I_3 is the last emerging intensity of light after it has pass through the polarizing stack

         Substituting values

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}

                           =234.622W/cm^2

When the second    is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

                      I_f_3 = \frac{I_p}{2} cos ^2 \theta_2

I_f_3 is the intensity of the light emerging from the stack

                     

substituting values

                     I_f_3 = \frac{234.622}{2} * cos^2(61)

                       I_f_3 = 27.57 W/cm^2

  When the third polarizer is removed  the  second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be  

                  I_f_2 = \frac{I_p}{2} cos ^2 \theta_1

I_f_2 is the intensity of the light emerging from the stack

Substituting values

                  I_f_2 =  \frac{234.622}{2} cos^2 (21)

                     I_f_2 = 102.24 W/cm^2

   

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2 years ago
Camille knows that range of motion is very important. She is designing a weekly exercise program and is not sure where flexibili
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The best activity for her to do to improve her range of motion is flexibility.

<h3>What are a few range of motion illustrations?</h3>

The term the range of motion (ROM) describes the extent to which a joint or muscle may be moved or stretched. Everybody has a distinct experience. For instance, whereas some people can perform a complete split, others cannot because their joints are stiff and their muscles are unable to extend as far.

<h3>What restricts motion range?</h3>

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To knoiw more about range of motion visit:

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1 year ago
A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W
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Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, P_{gauge \ pressure} = 2.95 atm

Atmospheric pressure, P_{atm} = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2

where;

P₁ = P_{gauge \ pressure} + P_{atm \ pressure}

P₂ = P_{atm}

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 =  P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 =  P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 =  P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} +  \rho gz_1 =  \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} +  \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} +  \rho gz_1)}{\rho} }

where;

\rho is the density of seawater = 1030 kg/m³

v_2 = \sqrt{ \frac{2(2.95*101325 \ + \  1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s

Therefore, the water is flowing at the rate of 28.04 m/s.

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Answer:

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