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ryzh [129]
2 years ago
15

You are standing next to a table and looking down on a record player sitting on the table. Take the spindle (axis of rotation) t

o be the center of your coordinate system and the y axis to be perpendicular to the side of the player you are standing next to. Long-playing records revolve 33(1/3) times per minute. You put a small blob of clay at the edge of a record that has a radius of 0.15 m, positioning the clay such that it is at its greatest value of y at t = 0.
Equation of motion for the y component of the clay's position: y(t)=Asin(ωt+ϕi)

Required:
a. What is the rotational speed of the clay?
b. Determine the value of A in the equation of motion.
c. Determine the value of ϕi in the equation of motion. Suppose that −π<ϕi≤π
Physics
1 answer:
VLD [36.1K]2 years ago
7 0

Answer:

a) the rotational speed of the clay is  3.45 rad/s

b) the value of A in the equation of motion is 0.15 m

c) the value of ϕi is 90° or π/2 rad.

Explanation:

 Given that;

Revolution per minute rpm = 33( 1/3) =  100/3

The frequency f = 100 / 3(60) = 0.55 Hz

a)

Rotational speed W = 2πf

we substitute

W = 2π × 0.55

W = 3.45 rad/s

Therefore, the rotational speed of the clay is  3.45 rad/s

b)

given equation; y(t)=Asin(ωt+ϕi)

given that radius = 0.15 m

y(t)=(0.2)sin(ωt+ϕi)

Therefore, the value of A in the equation of motion is 0.15 m

c)

since y(t) has the maximum value at t =0

so at t=0

y(0) = (0.15)sin(ω(0)+ϕi)

= 0.15sin(ϕi)

this will give maximum value when ϕi = 90°

so

y(0) = (0.15)sin(ω(0)+ϕi)

= 0.15sin(90°)

= 0.15

hence, the value of ϕi is 90° or π/2 rad.

You might be interested in
You are riding in an elevator on the way to the
timofeeve [1]

Answer:

F = 104.832 N

Explanation:

given,

upward acceleration of the lift = 1.90 m/s²

mass of box containing new computer = 28 kg.

coefficient of friction = 0.32

magnitude of force = ?

box is moving at constant speed hence acceleration will be zero.

Now force acting due to lift moving upward =

               F = μ m ( g + a )

               F = 0.32 × 28 × ( 9.8 + 1.9 )

              F = 104.832 N

hence, the force applied should be equal to 104.832 N

4 0
3 years ago
Two blocks connected by a light string are being pulled across a frictionless horizontal tabletop by a hanging 16.2-N weight (bl
Artemon [7]

Newton's second law allows us to find the results for the string tensions are:

  • T₁ = 6.7 N
  • T₂ = 16.54 N

Newton's second law gives a relationship between force, mass and acceleration of bodies

            ∑ F = ma

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration.

Free-body diagrams are representations of the forces applied to bodies without the details of them.

The reference system is a coordinate system with respect to which the forces decompose, in this case the x-axis is parallel to the plane and the positive direction in the direction of movement, the y-axis is perpendicular to the plane.

In the attachment we see a free-body diagram of the three-block system.

Let's apply Newton's second law to each body.

Block C

Y-axis

       W_c -T_2 = m_c a

Block A

X axis  

       T_2 - T_1 - W_a_x = m_a a  

Y axis  

       N_a - W_a_y = 0  

Block B

X axis

      T_1 - W_b_x = m_b a  

Y axis

      N_b - W_b_y =0

Let's  use trigonometry to find the components of the weight.

Block A

         cos θ = \frac{W_a_y}{W_a}  

         sin θ = \frac{W_a_x}{W_a}

         W_a_y = W_a cos \theta

         W_a_x= W_a sin \theta

Block B

        cos θ = \frac{W_b_y}{W_b}

        sin θ =  \frac{W_b_x}{W_b}

        W_b_y = W_b cos \theta \\W_b_x = W_b sin \theta

Let's write our system of equations.

     W_c - T_2 = m_c a \\           T_2 - T_1 - W_a_x = m_a a \\T_1 - W_b_x = m_b a

 

Let's find the acceleration of the bodies, adding the equations.

     W_c - W_a_x - W_b_x = ( m_a+m_b+m_c) a\\  

         

The weight is

    W = mg

Let's  substitute

         (m_c - m_a -m_b ) g \ sin \theta = ( m_c+m_a+m_b)  \ a  \\a= \frac{ m_c-m_a-m_b  }{ m_a+m_b+m_c} \ g sin \theta

Indicate ma mass of the block a ma = 1.00 kg, the mass of the block b mb = 2.2 kg and the weight of the block c Wc = 16.2 N, let's find the mass of block c.

          m_c = Wc / g

          m_c = 16.2 / 9.8

          m_c = 1.65 kg

we substitute the values

          a= \frac{1.65 -2.20 -1.00}{1.65+2.20+1.00} \ 9.8 \ sin \theta  \\a= -0.3096 sin \theta

The negative sign indicates that the system is descending, to be able to give a specified value an angle is needed, they assume that the angle of the ramp is 45º

          a = - 0.3196 sin 45

          a = -0.226 m / s

Taking the acceleration we are going to look for the tensions.

From the equation of block C

           W_c - T_2 = m_c a \\T_2 = m_c ( g-a)\\T_2 = 1.65 ( 9.8 + 0.226)

            T₂ = 16.54 N

From the equation of block B

          T_1 - W_b_x = m_b a\\T_1 = m_b (a + g sin \theta)\\T_1 = 1.00 (-0.226 + 9.8 \ sin 45)

           T₁ = 6.7 N

In conclusion using Newton's second law we can find the results for the string tensions are:

  •  T₁ = 6.7 N
  •  T₂ = 16.54 N

Learn more here:  brainly.com/question/20575355

7 0
3 years ago
Discuss renewable and nonrenewable resources. Give at least two advantages and two disadvantages of each.
lukranit [14]

wind = renew. wind needs to blow...very large "wind farms". no pollution

oil burns for heat or petrol. v high pollution. technology advanced

5 0
3 years ago
How much energy is need to raise 50 kg of water from 45 c to 80c?
Dmitriy789 [7]
Based on your problem, what you are looking for is the quantity of heat. To solve for it, you will need this formula:

Q = mc(T2-T1)

Where: Q = Quantity of heat
             m =  mass of the substance
             c  = Specific heat
             T2 = Final temperature
             T1 = Initial temperature

Now the specific heat of water is 4.184 J/(g°C), meaning that is how much energy is required to raise the temperature of 1g of liquid water by 1 degree Celsius. 

Since your mass is in kilograms, let us convert that into grams, which will be equal to 50,000 grams. Now we can put our given into the equation:

Q = mc(T2-T1)
   = 50,000g x  4.184 J/(g°C) x (80°C - 45°C)
   = 50,000 g x 4.184 J/(g°C) x 35°C   
   = 7,322,000 J or 7,322 kJ or 7.322 MJ

8 0
3 years ago
Read 2 more answers
Please help ASAP. A car engine with a mass of 1750 kg can exert 215 000 W. a) How long will it take the car to reach 95.0 km/h i
tensa zangetsu [6.8K]
<h2>Answer:</h2><h3><u>QUESTION①)</u></h3>

<em>✔ First step : calculate the kinetic energy that this car requires to reach 95 km/h</em>

95/ 3,6 ≈ 26,4 m/s  

<em>Ec = ½ m x V²  </em>

With Ec in J; m in kg; and V in m/s  

  • Ec = ½ 1750 x 26,4²  
  • Ec ≈ 610 000 J

<em>✔ Knowing that the car has a p power of 215,000 W, so : </em>

T = E/P  

  • T = 610 000/215 000  
  • T ≈ 2.8 s
<h3>The car takes 2.8 s to reach 95 km/h </h3>

<h3><u>QUESTION②)</u></h3>

N = 2,8/6,5 x 100 = 43.07

<h3>The car efficiency is 43 % </h3>
5 0
3 years ago
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