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Mazyrski [523]
3 years ago
6

Will an object with a density of 1.05 g/ml float or sink in water? Explain.

Physics
1 answer:
WINSTONCH [101]3 years ago
7 0
It will sink because it is heavier. The density of water 1.00 g/ml
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Find equivalent resistance between A and B​
V125BC [204]

Answer:

In the picture

Explanation:

I hope that it's a clear solution and explanation, hope that helps.

8 0
3 years ago
Any one tell me about the earth rotation it rotatining or not with any proof? ​
Dafna11 [192]
The proof that the earth is rotating is the happens of night and day also the seasons, eg. winter, summer, autumn.
6 0
3 years ago
A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the
vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table.  R = 0.287 kJ/kg-k

The Pressure-Volume relation is <em>PV</em> = <em>C</em>

<em>T = C </em> for isothermal process

Calculating for the work done in isothermal process

<em>W</em> = <em>P</em>₁<em>V</em>₁ ln[\frac{P_{1} }{P_{2} }]

   = <em>mRT</em>₁ln[\frac{P_{1} }{P_{2} }]      [∵<em>pV</em> = <em>mRT</em>]

   = (5) (0.287) (272.039) ln[\frac{2.0}{1.0}]

   = 270.588 kJ

Since the process is isothermal, Internal energy change is zero

Δ<em>U</em> = mc_{v}(T_{2}  - T_{1} ) = 0

From 1st law of thermodynamics

Q = Δ<em>U  </em>+ <em>W</em>

   = 0 + 270.588

   = 270.588 kJ

4 0
3 years ago
A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo
CaHeK987 [17]

Answer:\omega =93.51 rad/s

Explanation:

Given

mass of Flywheel m_1=1440 kg

mass of bus m_b=10200 kg

radius of Flywheel r=0.63 m

final speed of bus v=21 m/s

Conserving Energy i.e.

0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus

Let \omegabe the angular velocity of Flywheel

0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}

I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2

0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}

\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}

\omega =21\times 4.45=93.51 rad/s

8 0
3 years ago
Although it shouldn’t have happened, on a dive i fail to watch my spg and run out of air. if my buddy is close by, my best optio
leva [86]

Answer:

B ) Ascend using my buddy alternative air source / make an emergency Ascent

Explanation:

From the description it can be seen his buddy is close by of which he can easily use the alternative air source. Also we can see that he is closer to the water surface than his buddy, of which controlled emergency swimming ascent is highly favourable in this condition.

5 0
3 years ago
Read 2 more answers
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