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Lelechka [254]
3 years ago
14

A 110-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a

bout the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s
Physics
1 answer:
sweet [91]3 years ago
8 0

Answer:

155 N

Explanation:

We are given following data for a uniform, solid, horizontal disk:  

r = 1.5 m

m= 110 kg

t=2 s

w = 0.6  

Torque is given by:  

т = F*r

  =I*α

Solving it for the force exerted on the rope:  

F = I*α/r

  = (1/2*m*r^2)*(2π*w/t )/r

  = (1/2*110*1.5^2)*(2π*0.6/2 )/1.5

  = 155 N

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3 years ago
A coin is placed on a large disk which rotates uniformly at a rate of 1 rot/s. The coefficient of friction between the coin and
s2008m [1.1K]

Answer:

r = 0.02 m

Explanation:

from the question we have :

speed = 1 rps = 1x 60 = 60 rpm

coefficient of friction (μ) = 0.1

acceleration due to gravity (g) = 9.8 m/s^{2}

maximum distance without falling off (r) = ?

to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force

mv^2 / r =  m x g x μ

v^2 / r =  g x μ  .......equation 1

where

velocity (v) = angular speed (rads/seconds) x radius

angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm

angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds

now

velocity = 6.28 x r = 6.28 r

now substituting the value of velocity into equation 1

v^2 / r =  g x μ

(6.28r)^2 / r = 9.8 x 0.1

39.5 x r = 0.98

r = 0.02 m

6 0
3 years ago
express each of the following aa indicated a. 2 dm expressed in millimeter b. 2h 10min expessed in sec c. 16g expressed in centi
Alex787 [66]
I thinks or know it is A or B cuz magnets and metal attract
4 0
3 years ago
Question 7 (1 point)
ioda

Answer:

1150 secs

distance = speed x time

time = distance / speed

230,000 / 200 = 1150

time = 1150 seconds

7 0
3 years ago
during a typical afternoon thunderstorm in the summer, an area of 66.0 km2 receives 9.57 108 gal of rain in 18 min. how many inc
zheka24 [161]

Answer:

2.16 inch

Explanation:

area under water = 66 km²

= 66 x ( 3280.84 x 12 )² inch²

= 1.023 x 10¹¹ sq inch

volume of rain = 9.57 x 10⁸  gallon = 9.57 x 10⁸ x 231 inch³

= 2.21 x 10¹¹ inch³

If depth of rainfall be t

volume of rain = surface area x depth

= 1.023 x 10¹¹ x t

So ,

1.023 x 10¹¹ x t  = 2.21 x 10¹¹

t = 2.16 inch

5 0
2 years ago
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