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Genrish500 [490]
3 years ago
5

A 1.6 kg bicycle wheel with a radius of 0.31 m turns at a constant angular speed of 27 rad/s when a(n) 0.33 kg reflector is at a

distance of 0.19 m from the axle. What is the angular speed of the wheel when the reflector slides to a distance of 0.26 m from the axle? Answer in units of rad/s.
Physics
1 answer:
skad [1K]3 years ago
3 0

Answer:

angular speed = 25.393 rad/s

Explanation:

Given Data:

weight of bicycle wheel is 1.6 kg

radius of bicycle wheel is 0.31 m

angular speed is 27 rad/s

weight of reflector uis 0.33 kg

reflector distance from axle is 0.19 m

we know that moment of inertia between two mass is calculated as

Moment of inertiaI1 = \frac{1}{2} { MR^2 +m r^2 }

= \frac{1}{2}[ {1.6*0.31^2 + 0.33* 0.19^2}]

= 0.0828

Moment of inertia I2

= \frac{1}{2}[ {1.6*0.31^2 + 0.33* 0.26^2}]

= 0.088

\frac{I1}{I2} = 0.9405

\omega =  {\frac{I1}{I2} \omega _1

= 0.9405 \times 27

             = 25.393 rad/s

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Elan Coil [88]

Solution:

Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg

Temperature = 200 °C

And 25% of the volume by liquid water is steam.

State 1

$m=\frac{V}{v}$

$m=m_f+m_g$

$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$

$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$       (taking the value of $v_g$ and $v_g$ at 200°C  )

$V=6.304 \times 10^{-3}$

Now quality of vapor

$x=\frac{m_g}{m}$

  $=3.377 \times 10^{-3}$

Internal energy at state 1 can be found out by

$u_1=u_f+xu_{fg}$

    $=850.65+3.377\times10^{-3}\times 1744.65$

    = 856.54 kJ/kg

After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor $v_2=v_g \text { and }\ x=1$

Tank is rigid, so volume of tank is constant.

$v_g=v_2=\frac{V}{m}$

$v_g=\frac{6.304\times 10^{-3}}{1.4}$

$v_g=4.502 \times 10^{-3} \ m^3 /kg$

Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature

$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$

   = 369.11° C

Internal energy at state 2

$u_2=2154.9 \ kJ/kg$

Now power rating of the resistor

$P=\frac{m(u_2-u_1)}{t}$

$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$

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5 0
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Which statement about polymers is true?
Mamont248 [21]
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6 0
3 years ago
A car of mass 2400kg moving at 20 m/s slams into a cement wall and comes to a halt?
Brut [27]

Impulse = change in momentum

The car's momentum was (mass) x (speed)

Momentum = (2400 kg) x (20 m/s)

Momentum = 48,000 km-m/s

To completely stop the car, the impulse = -48,000 km-m/s .

7 0
3 years ago
If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t
solong [7]
Refer to the diagram shown below.

Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.

The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m

The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²

The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
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Answer: 1.375 s

3 0
3 years ago
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