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vitfil [10]
3 years ago
12

The flywheel of a steam engine runs with a constant angular speed of 113 $rev/min$. When steam is shut off, the friction of the

bearings and the air brings the wheel to rest in 1.0 $h$. What is the magnitude of the constant angular acceleration of the wheel in $rev/min^2$? Do not enter the units.
Physics
1 answer:
Oksana_A [137]3 years ago
6 0

Answer:

α = - 1.883 rev/min²

Explanation:

Given

ωin = 113 rev/min

ωfin = 0 rev/min

t = 1.0 h = 60 min

α = ?

we can use the following equation

ωfin = ωin + α*t      ⇒     α = (ωfin - ωin) / t

⇒     α = (0 rev/min - 113 rev/min) / (60 min)

⇒     α = - 1.883 rev/min²

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A stone is dropped into a well. The sound of the splash is heard 3.5 seconds later. What is the depth of the well? Take the spee
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Answer:

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

s = ut +\frac{1}{2}at^{2}

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

s = \frac{1}{2}gt'^{2}

s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}                     (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

s = v\times t''

s = 343\times t''                                              (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t'                                                                        (3)

Using eqn (2) and (3):

s = 343(3.5 - t')                                                                 (4)

from eqn (1) and (4):

4.9t'^{2} = 343(3.5 - t')

4.9t'^{2} + 343t' - 1200.5 = 0

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

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3 years ago
Compare the maximum rate of heat transfer to the basal metabolic rate by converting a bmr of 88 kcal/hr into watts. what is the
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Explanation :

It is given that,

BMR i.e basal metabolic rate is 88 kcal/hr. So, BMR in watts is converted by the following :

We know that, 1 kilocalorie = 4184 joules

So, 1\ kcal/h=\dfrac{1\times 4184\ J}{3600\ sec}

1\ kcal/h=1.16\ J/sec

J/sec is nothing but watts.

So, 1\ kcal/h=1.16\ watts

and 88\ kcal/h=88\times 1.16\ watts = 102.08\ watts

So, it can be seen that the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat.

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A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho
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Answer:

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Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f

There is no kinetic energy in the initial state, nor potential energy in the end,

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In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

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The inertia of the bodies is given by the equation,

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I=0.05Kgm^2

On the other hand the angular velocity is given by

\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s

Replacing these values in the equation,

(0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2

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h=2.86m

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