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krek1111 [17]
2 years ago
14

Find the time t1 it takes to accelerate the flywheel to ω1 if the angular acceleration is α. express your answer in terms of ω1

and α.
Physics
2 answers:
Ivahew [28]2 years ago
6 0

the time t1 it takes to accelerate:

\large{\boxed{\bold{t_1=\frac{\omega_1 }{\alpha}}}}

<h3>Further explanation </h3>

The speed of the rotating object is called angular velocity. This value is part of circular motion.

The angular velocity is denoted by w including the vector quantity. The unit used is usually rpm (rotation per minute) or radians / sec

Average angular velocity is the ratio of angular displacement to the time interval

\rm \omega=\dfrac{\Delta \theta}{\Delta t}

Circular motion with constant acceleration (angular acceleration denoted by α) can be formulated:

\rm \omega=\omega o+\alpha t

ωo = initial angular velocity

ω = angular velocity after t

 the flywheel accelerate to ω1, so:

wo = 0 because The flywheel is assumed to be at rest at time t = 0

so the equation becomes:

ω1 = ωo + αt

ω1 = 0 + αt

ω1 = αt

\large{\boxed{\bold{t=\frac{ \omega_1}{\alpha }}}

Learn more

the average velocity

brainly.com/question/5248528

resultant velocity

brainly.com/question/4945130

velocity position

brainly.com/question/2005478

VMariaS [17]2 years ago
3 0
T1 = ω1/α.............
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aniked [119]

Answer:

a. N = 2.49W b.  0.40

Explanation:

a. What is the magnitude of the normal force FNFN between a rider and the wall, expressed in terms of the rider's weight W?

Since the normal force equals the centripetal force on the rider, N = mrω² where r = radius of cylinder = 3.05 m and ω = angular speed of cylinder = 0.450 rotations/s = 0.450 × 2π rad/s = 2.83 rad/s

Now N = mrω² = m(3.05 m) × (2.83 rad/s)² = 24.43m

The rider's weight W = mg = 9.8m

The ratio of the normal force to the rider's weight is

N/W = 24.43m/9.8m = 2.49

So the normal force expressed in term's of the rider's weight is

N = 2.49W

b. What is the minimum coefficient of static friction µsμs required between the rider and the wall in order for the rider to be held in place without sliding down?

The frictional force, F on the rider by the wall of the cylinder equals the weight, W of the rider. F = W.

Since the frictional force F = μN, where μ = coefficient of static friction between rider and wall of cylinder and N = normal force between rider and wall of cylinder.

So, the normal force equals

N = F/μ = W/μ = mg/μ = mrω²

μ  = mg/mrω²

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6 0
2 years ago
An ideal monatomic gas initially has a temperature of T and a pressure of p. It is to expand from volume V1 to volume V2. If the
yawa3891 [41]

Answer:

Isothermal :   P2 = ( P1V1 / V2 ) ,  work-done pdv = nRT * In( \frac{V2}{v1} )

Adiabatic : : P2 = \frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }  , work-done =

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Explanation:

initial temperature : T

Pressure : P

initial volume : V1

Final volume : V2

A) If expansion was isothermal calculate final pressure and work-done

we use the gas laws

= PIVI = P2V2

Hence : P2 = ( P1V1 / V2 )

work-done :

pdv = nRT * In( \frac{V2}{v1} )

B) If the expansion was Adiabatic show the Final pressure and work-done

final pressure

P1V1^y = P2V2^y

where y = 5/3

hence : P2 = \frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }

Work-done

W = (3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)

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