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Ostrovityanka [42]
3 years ago
15

You are working as a letter sorter in a U.S. Post Office. Postal regulations require that employees' footwear must have a minimu

m coefficient of static friction of 0.5 on a specified tile surface. You are wearing athletic shoes for which you do not know the coefficient of static friction. In order to determine the coefficient, you imagine that there is an emergency and start running across the room. You have a coworker time you, and find that you can begin at rest and move 4.38 m in 1.21 s. If you try to move faster than this, your feet slip. Assuming your acceleration is constant, does your footwear qualify for the postal regulation?
Physics
1 answer:
Serggg [28]3 years ago
7 0

Answer:

  μ = 0.66,    therefore if it compiesy with the regulations

Explanation:

Let's solve this exercise in part, let's start by finding with kinematics the acceleration of man

           y = v₀ t + ½ a t²

as it starts from rest the initial velocity is zero

          y = ½ a t²

          a = \frac{2y}{t^2}

          a =\frac{2 \ 4.38}{1.21^2}  

          a = 6.46 m / s²

Now let's use Newton's second law,

Axis y

         N- W = 0

         N = W

         N = m g

X axis

on this axis the man exerts a backward force and by the law of action and reaction the floor exerts a forward force of the same magnitude, this forward force is the friction force.

         fr = m a

     

the friction force has an expression

          fr = my N

let's substitute

      μ mg = m a

      μ = a / g

   

let's calculate

     μ = 6.46 / 9.8

     μ = 0.66

therefore if you comply with the regulations

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Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

T=2\pi\sqrt{\frac{I}{K}}

I=mR^{2}

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}

As dimensions of new balance wheel are one-third of their original values

R_{new}=\frac{R}{3}

\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega

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1) Determine the magnitude of energy for each of the blanks on the diagram. Give the correct values for 1A, 1B, and 1C.
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Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa
tester [92]

Answer:

Part a)

t = 2.83 s

Part b)

Ball thrown downwards =v_f = 23.8 m/s

Ball thrown upwards =v_f = 23.8 m/s

Part c)

d = 22.24 m

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

\Delta y = 0 = v_y t + \frac{1}{2}at^2

0 = 13.9 t - \frac{1}{2}(9.81) t^2

t = 2.83 s

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

v_f^2 - v_i^2 = 2 a \Delta y

v_f^2 - (13.9)^2 = (2)(-9.81)(-19.1)

v_f^2 = 567.9

v_f = 23.8 m/s

Part c)

Relative speed of two balls is given as

v_{12} = v_1 - v_2

v_{12} = (13.9) - (-13.9) = 27.8 m/s

now the distance between two balls in 0.8 s is given as

d = v_{12} t

d = 27.8 \times 0.8

d = 22.24 m

7 0
4 years ago
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