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madreJ [45]
3 years ago
5

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f = 2.20 Hz. On this tray is an empty cu

p. Obtain the coefficient of static friction between the tray and the cup, given that the cup begins slipping when the amplitude of the motion is 5.50 10-2 m.
Physics
1 answer:
Kruka [31]3 years ago
6 0

Answer:

Coefficient of friction will be 0.9738

Explanation:

We have given frequency of the motion f = 2.20 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 2.20=13.816rad/sec

Amplitude of the motion A=5.50\times 10^{-2}m

Maximum acceleration is given by a=\omega ^2A=13.816^2\times 5.50\times 10^{-2}=9.544rad/sec^2

Acceleration due to gravity g=9.8m/sec^2

We know that acceleration is given by a=\mu g

\mu =\frac{a}{g}=\frac{9.544}{9.8}=0.9738

So coefficient of friction will be 0.9738

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Marizza181 [45]
The answer:
1.8 x 10^0 V
7 0
3 years ago
A plane is traveling at 300 m/s. How far will it travelin 1 hour?
Airida [17]

Answer:

1080000

Explanation:

300 x 60s=18000m/minute

18000 x 60min=1080000m/h

Explanation:

3 0
3 years ago
Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their
Vera_Pavlovna [14]

Answer:

A = 2.36m/s

B = 3.71m/s²

C = 29.61m/s2

Explanation:

First, we convert the diameter of the ride from ft to m

10ft = 3m

Speed of the rider is the

v = circumference of the circle divided by time of rotation

v = [2π(D/2)]/T

v = [2π(3/2)]/4

v = 3π/4

v = 2.36m/s

Radial acceleration can also be found as a = v²/r

Where v = speed of the rider

r = radius of the ride

a = 2.36²/1.5

a = 3.71m/s²

If the time of revolution is halved, then radial acceleration is

A = 4π²R/T²

A = (4 * π² * 3)/2²

A = 118.44/4

A = 29.61m/s²

7 0
3 years ago
A car with a mass of 2.0 * 10^3 kg is traveling at 15 m/s. what is the momentum of the car?
Allushta [10]
Hello,


Your answer to this problem is 400/3


Hope this helps!
3 0
3 years ago
A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in.-thick plate by welding along a helix that forms an angle of 25
Varvara68 [4.7K]

Answer:

Normal stress = 66/62.84 = 1.05kips/in²

shearing stress  = T/2 = 0.952/2 = 0.476 kips/in²

Explanation:

A steel pipe of 12-in. outer diameter  d₂ =12in  d₁= 12 -4in = 8in

4 -in.-thick  

angle of 25°

Axial force P = 66 kip axial force

determine the normal and shearing stresses

Normal stress б = force/area = P/A

           = 66/ (П* (d₂²-d₁²)/4

           =66/ (3.142* (12²-8²)/4

          = 66/62.84 = 1.05kips/in²

Tangential stress T = force* cos ∅/area = P/A

           = 66* cos 25/ (П* (d₂²-d₁²)/4

           =59.82/ (3.142* (12²-8²)/4

          = 59.82/62.84 = 0.952kips/in²

shearing stress  = tangential stress /2

                           = T/2 = 0.952/2 = 0.476 kips/in²

8 0
3 years ago
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