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luda_lava [24]
3 years ago
13

Particles q1, 92, and q3 are in a straight line.

Physics
1 answer:
dangina [55]3 years ago
6 0

Answer: 0

Explanation: Acellus

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Consider the following four objects: a hoop, a solid sphere, a flat disk, a hollow sphere. Each of the objects has mass M and ra
Svetlanka [38]

Answer:

Hoop.

Explanation:

The angular acceleration performed at a given torque:

\alpha = \frac{\tau}{I}

The moments of inertia of each element are described below:

Hoop

I = M\cdot R^{2}

Solid sphere

I = \frac{2}{5}\cdot M \cdot R^{2}

Flat disk

I = \frac{1}{2}\cdot M \cdot R^{2}

Hollow sphere

I = \frac{2}{3}\cdot M \cdot R^{2}

The greater the moment of inertia, the greater the torque to obtain the same angular acceleration. Therefore, the hoop requires the largest torque to receive the same angular acceleration.

8 0
4 years ago
A bicycle rider travels 50.0 Km in 2.5 hours. What is the cyclist's average speed?
vekshin1
<span>If you divide the numbers you will get on average of 20km</span>
5 0
3 years ago
Read 2 more answers
What is the nucleus of the atom
luda_lava [24]

Answer:

An atomic Nucleus

Explanation:

4 0
3 years ago
Read 2 more answers
200-grams of computer chips with a specific heat of 0.3 kJ/kg·K are initially at 25°C. These chips are cooled by placement in 0.
balu736 [363]

Answer:

a. -0.01324 kJ/K,  b.  = 0.03233 kJ/K , c.  = 0.01909, Yes the process is possible

Explanation:

Heat transfer will occur between the chip and the surrounding fluid. Then, finally they will attain a common equilibrium temperature and heat transfer will stop. Now, if we assume that, after heat transfer, chip will attain the temperature of fluid, that is, -34 C,, So , to check whether this is possible

Amount of energy lost by the chip = m . c . (T(i) - T(f))

= 0.2 x 0.3 (25 + 34) = 3.54 KJ

Now, to evaluate the final state of the fluid, after the heat transfer completion,

Energy Gained = m(mew final – mew initial) = m[(μf+ x . μfg) - μf]

Note that heat transfer will change the internal energy of the fluid. Do not consider enthalpy change, as this is not a problem involving fluid flow in and out of the system

M[(μf+ x . μfg) - μf] = m(xμfg)

<u>Energy gained by the fluid will be equal to the energy lost by the chip (No energy loss to the surroundings)</u>

3.54 = 0.1 . X x 203.29

<u>x = 0.1741, which is the dryness fraction of fluid at the final state.</u>

Observe that the total energy lost by the chips is 3.45 kJ and fluid R-134a has got its value of mew fg at -34 C which is = 203.29 kJ/kg

So for 0.1kg of R-134a

0.1 x μfg= <u>20.329 kJ, which is much greater than 3.45 kJ</u>, therefore, it is certain that the state of fluid will be at -34 C only and at the saturation pressure of 69.56 KPa. So the chip will come to attain the temperature of -34 C.  

a. Write the equation for the change of entropy in the chips

ΔSchips = mchips . c . ln(T2/T1), where mc is the mass of chips, c is the specific heat of chips, T2 is the temperature at state 2 and T1 is the temperature at state 1

Substitute mc = 0.2 kg, c = 0.3kJ/kg.K, T1 = 25 + 273, T2 = -34 + 273

delSchips = 0.2 x 0.3 x ln [(-34+273)/ (25+273)]

= -0.01324 kJ/K

There fore the change in entropy of the chips is -0.01324 kJ/K

b. Entropy change of fluid R- 134a

ΔS2 = m[Sfinal – S initial]

= m[Sf + x . Sfg - Sf]

= 0.2 x (0.1741 x 0.92859)

= 0.03233 kJ/K

c. Calculate the total change in the entropy of the entire system

delS = delSchips + delSR -134a

= -0.01324 + 0.03233

= 0.01909

<u>Since the total change in entropy of the entire system is positive that exactly explains that the actual processes are happening in the direction of increase of entropy therefore, the process is possible.</u>

<u />

6 0
3 years ago
A chamber fitted with a piston contains 1.90 mol of an ideal gas. Part A The piston is slowly moved to decrease the chamber volu
yarga [219]

Answer:

The work done is 5136.88 J.

Explanation:

Given that,

n = 1.90 mol

Temperature = 296 K

If the initial volume is V then the final volume will be V/3.

We need to calculate the work done

Using formula of work done

W=nRT\ ln(\dfrac{V_{f}}{V_{i}})

Put the value into the formula

W=1.90\times8.314\times296\ ln(\dfrac{\dfrac{V}{3}}{V})

W=1.90\times8.314\times296\ ln(\dfrac{1}{3})

W=−5136.88\ J

The Work done on the system.

Hence, The work done is 5136.88 J.

5 0
3 years ago
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