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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.

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Answer:

yup

Explanation:

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3 years ago
A hula hoop is rolling along the ground with a translational speed of 26 ft/s. It rolls up a hill that is 16 ft high. Determine
nikklg [1K]

Answer:12.8 ft/s

Explanation:

Given

Speed of hoop v=26\ ft/s

height of top h=16\ ft

Initial energy at bottom is

E_b=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2

Where m=mass of hoop

I=moment of inertia of hoop

\omega=angular velocity

for pure rolling v=\omega R

I=mR^2

E_b=\frac{1}{2}mv^2+\frac{1}{2}mR^2\times (\frac{v}{R})^2

E_b=mv^2=m(26)^2=676m

Energy required to reach at top

E_T=mgh=m\times 32.2\times 16

E_T=512.2m

Thus 512.2 m is converted energy is spent to raise the potential energy of hoop and remaining is in the form of kinetic and rotational energy

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Therefore

163.8 m=mv^2

v=\sqrt{163.8}

v=12.798\approx 12.8\ ft/s

7 0
3 years ago
A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with
Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\  g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s

\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM

Therefore the speed in RPM will be 62.64 RPM.

5 0
3 years ago
The total electric flux from a cubical box 26.0 cm on a side is 1840 N m2/C. What charge is enclosed by the box?
vitfil [10]

The expression of the electric flux is

\Phi = \frac{Q}{\epsilon_0}

Here,

Q = Total charge enclosed in the closed surface

\epsilon_0 = Permittivity due to free space

Rearranging to find the charge,

Q = \epsilon_0 \Phi

Replacing with our values we have finally

Q = (8.85*10^{-12}F\cdot m^{-1})(1.84*10^3 N\cdot m^2/C)

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The charge enclosed by the box is 0.1684nC

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7 0
3 years ago
A car of mass 1600 kg can just be lifted what is the least force that the electronicmagnet must use to lift the car ?
Mkey [24]

The car's mass is 1600 kg.

Its weight is (mass) x (gravity).  

On Earth, that's (1600 kg) x (9.8 m/s²)  =  15,680 Newtons.

At the moment, that's the only force acting on the car, directed downward and provided by gravity.

If you want to lift the car, then the net force has to be directed upward, and must either exactly cancel or exceed the force of gravity.

So the minimum force required to lift the car is <em>15,680 Newtons</em>, directed vertically upward.

5 0
3 years ago
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