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myrzilka [38]
2 years ago
8

Some special vehicles have spinning disks (flywheels) to store energy while they roll downhill. They use that stored energy to l

ift themselves uphill later on. Their flywheels have relatively small rotational masses but spin at enormous angular speeds. How would a flywheel’s kinetic energy change if its rotational mass were 7 times larger but its angular speed were 7 times smaller?
Physics
1 answer:
Evgesh-ka [11]2 years ago
4 0

Answer:

Smaller by 7 times

Explanation:

Rotational mass is called moment of inertia

So, initial moment of inertia, I1 = I

initial angular velocity, ω1 = ω

Final moment of inertia, I2 = 7I

Final angular velocity, ω2 = ω/7

The kinetic energy in rotational motion is given by

K = \frac{1}{2}I\omega ^{2}

So, initial kinetic energy of rotation

K_{1} = \frac{1}{2}I_{1}\omega_{1} ^{2}= \frac{1}{2}I\omega ^{2}

So, final kinetic energy of rotation

K_{2} = \frac{1}{2}I_{2}\omega_{2} ^{2}= \frac{1}{2}7I \left (\frac{\omega}{7}  \right ) ^{2}

K_{2} = \frac{K_{1}}{7}

Thus, teh kinetic energy becomes smaller by 7 times.

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If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it
Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of  27.0 m/s2  is

      \frac{\Delta m}{\Delta t}   = 83.92 \ Kg/s

Explanation:

From the question we are told that

   The mass of the rocket is m = 6300 kg

   The velocity at gas is being ejected is  u =  2000 m/s

    The initial acceleration desired is a =  27.0 \  m/s

   The time taken for  the gas to be ejected is  t = 1 s

Generally this desired acceleration is mathematically represented as

        a = \frac{u *  \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}

Here \frac{\Delta m}{\Delta  t }  is the rate at which gas is being ejected with respect to time

Substituting values

      27 = \frac{2000 *  \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}

=>   170100 -27* \frac{\Delta m}{\Delta t} = 2000 *  \frac{\Delta m}{\Delta t}

=>   170100  = 2027 *  \frac{\Delta m}{\Delta t}

=>   \frac{\Delta m}{\Delta t}   = \frac{170100}{2027}

=>   \frac{\Delta m}{\Delta t}   = 83.92 \ Kg/s

     

3 0
2 years ago
The late news reports the story of a shooting in the city. Investigators think that they have recovered the weapon and they run
Gemiola [76]

Answer:

50000 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) of bullet = 0.050 kg

velocity (v) = 400 m/s

Distance (s) = 0.080 m

Force (F) =?

Next, we shall determine the acceleration of the bullet. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 400 m/s

Distance (s) = 0.080 m

Acceleration (a) =?

v² = u² + 2as

400² = 0 + (2 × a × 0.08)

160000 = 0 + 0.16a

160000 = 0.16a

Divide both side by 0.16

a = 160000 / 0.16

a = 1×10⁶ m/s²

Finally, we shall determine the force exerted by the bullet on the target. This can be obtained as follow:

Mass (m) of bullet = 0.050 kg

Acceleration (a) of bullet = 1×10⁶ m/s²

Force (F) =?

F = ma

F = 0.050 × 1×10⁶

F = 50000 N

Thus, the bullet exerted a force of 50000 N on the target.

7 0
2 years ago
Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exi
PIT_PIT [208]

Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

A.

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

B.

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

= 51.37455 kW

= 51.38 kW.

5 0
2 years ago
A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the
krek1111 [17]

Answer:

L = mp*v₀*(ms*D) / (ms + mp)

Explanation:

Given info

ms = mass of the hockey stick

uis = 0 (initial speed of the hockey stick before the collision)

xis = D (initial position of center of mass of the hockey stick before the collision)

mp = mass of the puck

uip = v₀ (initial speed of the puck before the collision)

xip = 0 (initial position of center of mass of the puck before the collision)

If we apply

Ycm = (ms*xis + mp*xip) / (ms + mp)

⇒  Ycm = (ms*D + mp*0) / (ms + mp)

⇒  Ycm = (ms*D) / (ms + mp)

Now, we can apply the equation

L = m*v*R

where m = mp

v = v₀

R = Ycm

then we have

L = mp*v₀*(ms*D) / (ms + mp)

5 0
2 years ago
"As the Voyager spacecraft penetrated into the outer solar system, the illumination from the Sun declined. Relative to the situa
____ [38]

Answer:

\frac{I_{2}}{I_{1}} = 0.04

Explanation:

The intensity of a star noticed at a certain distance is inversely proportional to the square of distance. Then:

I_{1}\cdot r_{1}^{2} = I_{2}\cdot r_{2}^{2}

The intensity of the Sun in Jupiter relative to Earth is:

\frac{I_{2}}{I_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}}

\frac{I_{2}}{I_{1}} = \left(\frac{1\,AU}{5.2\,AU} \right) ^{2}

\frac{I_{2}}{I_{1}} = 0.04

3 0
2 years ago
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