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larisa86 [58]
3 years ago
12

A turntable has an angular velocity of 3.5 rad/s. A dust bunny is on the disk of the turntable at a distance of 0.2m from the ce

nter of rotation. The coefficient of static friction required to keep the dust bunny from getting slung off is at leastA. 0.51B. 0.45C. 0.33D. 0.12E. 0.25
Physics
1 answer:
diamong [38]3 years ago
3 0

Answer:

E. 0.25

Explanation:

Given that

Angular speed ω = 3.5 rad/s

Distance ,r= 0.2 m

Lets take mass of dust bunny = m

We know that

Radial force F = m ω² r

The friction force on the dust bunny Fr

Fr= μ m g

To getting slung off  dust bunny from disk

m ω² r =  μ m g

ω² r =  μ  g

3.5²  x 0.2 =  μ x 10                        ( take g =10 m/s²)

μ = 0.245

μ = 0.25

Therefore answer is E

E. 0.25

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Calculate the elastic potential energy stored in a spring if it has a force constant of 150 N/m. the spring is extended to a len
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Answer:

6.75J

Explanation:

U=1/2KΔx²

U=0.5* 150*0.30^2

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2 years ago
When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1lbm=0.453
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Answer:

a) 0.022%

b) 10014.32 lb

Explanation:

a) Percentage uncertainty would be

0.0001\times \frac{100}{0.4539}=0.022%

Percent uncertainty is 0.022%

b) For 1 kg uncertainty mass in kg would be

\frac{1}{0.022}\times {100}=4545.5\ kg

Mass in pounds would be

\frac{4545.5}{0.4539}=10014.32\ lb

Mass in pound-mass is 10014.32 lb

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3 years ago
Compared to Earth's moon, the moons of Mars are
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I think is b I think
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2 years ago
3. What is the difference between Synthetic Oil, Semi-Syntheic oil (Blend), High milage motor oil, and conventional motor oil? ​
MariettaO [177]

Answer:

When evaluating synthetic blends, it's helpful to define the terms “synthetic blend” and “semi-synthetic”. Generally speaking, synthetic blends and semi-synthetic refer to the same thing: an oil that uses a combination of conventional and synthetic base oils in its formulation.

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6 0
3 years ago
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
kap26 [50]

Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

3 0
3 years ago
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