For this questions you  will need some mathematics and knowledge of engineering thermodynamics. Appropriate links are provided at the bottom of this section. As we saw previously all Carnot heat engine have the same efficiency irrespective of the working gas we are using. Hence, we might as well use an ideal gas with temperature-independent specific heats because its properties are precisely known and can be represented in terms of fairly easy equations. All noble gases ( helium, neon, argon, krypton, xenon, and the radioactive radon ) come extremely close to being an ideal gas at temperatures and pressure relevant to Stirling engine.

(6)       p V = m R_s T         Ideal gas law

(7)       Δu = c_v ΔT         Change of internal energy is proportional to change in temperature

<th>Figure 5 : Carnot Cycle</th>

1 → 2 : isothermal compression. Because temperature is constant the internal energy of the gas does not change according to Eq.(2). According to Eq.(1) we also have <i>p V = m R Tc= constant</i>. This makes it possible to evaluate the work integral Wc =&int; p dV This results in the equation : (8)       Qc=Wc = m Rs Tc ln ( V1 / V2 ) 3 → 4 : isothermal expansion. This follows exactly the logic as the process 1 → 2. Hence, (9)       Qh=Wh = m Rs Th ln ( V4 / V3 ) 2 → 3 : adiabatic compression. With κ = cp/cv for such process : (10)       V2/V3 = (Th/Tc)1/(κ-1) 4 → 1 : adiabatic expansion. Follows the same logic as process 2 → 3 with the result: (11)       V1/V4 = (Th/Tc)1/(κ-1)

We use Eq.(8) and (9) to find a first expression for the efficiency of the Carnot cycle (see also Eq.(3) ) :

(12)       η = 1 - Q_c/Q_h = 1 - (T_c ln ( V1 / V2 )) / (T_h ln ( V4 / V3 ))

From Eq. (10) and (11) we get :

(13)       V₁/V₂ = V₄/V₃

and therefore :

(14)       η = 1 - T_c/T_h

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Ans. $Let two angles of projection for which the horizontal range is same are \alpha and \beta .$

$Then the relation b/w them : \alpha + \beta = 90°$

 $Let two angles of projection for which the horizontal range is same are \alpha and \beta .Then the relation b/w them :\alpha +\beta = 90°$

$Ans . Proof : Let the velocity, pressure and area of a fluid column at a point X be v_1, p_{1}and A_1$

$and at another point Y be v_2, p_2 and A_2. Let the volume that is bounded by X and Y$

$be moved to M and N. let XM = L_1 and YN = L_2. Now if we can compress the fluid then we have,$

$A_1 \times L_1= A_2 \times L_2$

$We know that that the work done by the pressure difference per volume of the$

$unit is equal to the sum of the gain in kinetic energy and$

$gain in potential energy per volume of the unit$.

$This implies Work done = force \times dis\tan ce \Rightarrow Work done = p \times volume$

$Therefore, net work done per volume = p_1 – p_2$

$Also, kinetic energy per unit volume = \frac{1}{2} m{v}^2 = \frac{1}{2} \rho v^2$

$Therefore, we have,$

$Kinetic energy gained per volume of unit = \frac{1}{2} \rho \left({\left(v_2\right)}^2 - {\left(v_1\right)}^2\right)$

$And potential energy gained per volume of unit = pg (h_2 – h_1)$

$Here, h_1 and h_2 are heights of X and Y above the reference level taken in common.$

$Finally we have p_1 – p_2 = \frac{1}{2}\rho \left({\left(v_2\right)}^2-{\left(v_1\right)}^2\right) + \rho g (h_2 – h_1)$

$\Rightarrow p_1 + \frac{1}{2} \rho (v_1{)}^2 + \rho g{h}_1 = p_2 + \frac{1}{2}\rho (v_2{)}^2 + \rho g{h}_2$

$\Rightarrow p + \frac{1}{2}\rho v^2 + \rho gh is a cons\tan t When we have h_1 = h_2$

$Then we have, p + \frac{1}{2} \rho v^{2}is a cons\tan t. This proves the Bernoulli’s Theorem$

$Ans. According to Bernoulli’s theorem in physics, whenever there is an increase in the speed of the liquid, there is a simul\tan eous decrease in the potential energy of the fluid or we can say that there is a decrease in the pressure of the fluid. Basically, it is a principle of conservation of energy in the case of ideal fluids. If the fluid flows horizontally such that there is no change in the gravitational potential energy of the fluid then increase in velocity of the fluid results in a decrease in pressure of the fluid and vice versa.$

Ans. Lorentz Force is given by $\overrightarrow{F} = q\overrightarrow{v}\times \overrightarrow{B}$

as the force is always perpendicular to the motion and work done given by $W = F.S \cos \theta$, as angle is always 90 so work done will be Zero. 

Ans. a) The Average KE will be the Same as Conditions of temrature and pressure are same.

b) We know $V_{rms} = \sqrt{\frac{3pV}{M}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3RT}{mN}} = \sqrt{\frac{3kT}{m}}$

Where, M = Molar mass of the gas

m  = mass of each molecule of gas

R = gas constant 

k = Boltzmann constant

As Temprature is constant

So $V_{rms } \propto \sqrt{\frac{1}{m}}$

Given : $m_A>m_B>m_C$

So ${\left(V_{rms}\right)}_A<{\left(V_{rms}\right)}_B<{\left(V_{rms}\right)}_C$

=> ${\left(V_{rms}\right)}_C>{\left(V_{rms}\right)}_B>{\left(V_{rms}\right)}_A$

Ans. Viscosity is an internal property of a fluid that offers resistance to flow.

Ans. When there are no opposing forces, a moving body tends to keep moving with a steady velocity as we know from Newton's first law of motion. If, however, a resultant force does act on a moving body in the direction of its motion, then it will accelerate per Newton's second law $F = ma$ The work done by the force will become converted into increased kinetic energy in the body. 

Derivation Using Calculus :

Begin with the Work-Energy Theorem : The work that is done on an object is related to the change in its kinetic energy<meta aria-hidden="true" />

$∆K = W$

Rewrite work as an integral: we can represent the work done in terms of a velocity differential.

$∆K = \int F. dr$

Rewrite force in terms of velocity: mass is a scalar and can therefore be factored out.

$∆K = \int ma.dr$

=> $∆K = m\int a.dr$

=> $∆K = m\int \frac{dv}{dt}.dr$

=> $∆K = m\int \frac{dr}{dt}.dv$

=> $∆K = m\int v.dv$

=> $∆K = \frac{1}{2}m{v}^2$

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And. It's displacement

if I started from A and go to B in a direction and then returned back to A and go in a direction opposite to N then my velocity is -ve with respect to N.

A vector points in a direction in space. A negative vector (or more precisely "the negative of a vector") simply points the opposite way.

If I drive from my home to my workplace (and then defining my positive direction in that way), then my velocity is positive if I go to work, but negative when I go homefrom work. It is all about direction seen from how I defined my positive axis.

Ans. It is due to the internal frictional force that develops between different layers of fluids as they are forced to move relative to each other.

The reason behind frictional force between layers of fluids:

i) intermolecular force of cohesion
(ii) molecular momentum exchange

1. Stress is defined as a force that can cause a change in an object or a physical body while strain is the change in the form or shape of the object or physical body on which stress is applied.
2. Stress can occur without strain, but strain cannot occur with the absence of stress.
3. Stress can be measured and has a unit of measure while strain does not have any unit and, therefore, cannot be measured.
4. Strain is an object’s response to stress while stress is the force that can cause strain in an object

$P\alpha T$

Therfore, $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Substitute the initial and final values and find out the unknown component. 

generally in physics we all have observed that some values were taken 0 during some derivation and solving numericals.

Actually this step taken by us is the assumption taken by  us to simplify the complex terms whose solution is found to be difficult for even a common person to solve it.

we assume or compare only that angle of contact as 0 that is very very small on comparing with the other quantities which are very large on comparing with the angle. so in some numericals angle of contact is assumed to be zero

Two waves (with the same amplitude, frequency, and wavelength) are travelling in opposite directions. Using the principle of superposition, the resulting wave amplitude may be written as:

y ( x , t ) = y m sin ( kx - ωt ) + y m sin ( kx + ωt ) = 2 y m sin ( kx ) cos ( ωt )

This wave is no longer a travelling wave because the position and time dependence have been separated. The the wave amplitude as a function of position is <i>2y_m</i>sin(<i>kx</i>). This amplitude does not travel, but stands still and oscillates up and down according to cos(<i>ω t</i>). Characteristic of standing waves are locations with maximum displacement (antinodes) and locations with zero displacement (nodes)..

If two sinusoidal waves having the same frequency (and wavelength) and the same amplitude are travelling in opposite directions in the same medium then, using superposition, the net displacement of the medium is the sum of the two waves.

when the two waves are 180° out-of-phase with each other they cancel, and when they are exactly in-phase with each other they add together. As the two waves pass through each other, the net result alternates between zero and some maximum amplitude. However, this pattern simply oscillates; it does not travel to the right or the left, and thus it is called a "<i>standing wave</i>".

upper vale ne galat answer deya hai

To quantify elasticity, physics defines elasticity as "resistance to change". The greater the resistance to change, the greater is the elasticity of the material and the faster it comes back to its original shape or configuration when the deforming force is removed. By this definition, steel is more elastic than rubber because steel comes back to its original shape faster than rubber when the deforming forces are removed.

$\left(a\right) Since the spring potential energy is just enough to project the stone 10m high\Rightarrow P.E. of stone at 10m height = P.E. of the spring= mgh=(50/1000)*10*10=5J\left(b\right) Since 0.5*k*x^{2 }=m*g*\left(10\right) (when x= 3cm)Now 0.5*k*y^2 = m*g*\left(5\right)\Rightarrow y=x/\sqrt{2}\Rightarrow y=3/\sqrt{2}Hence it should be compressed by y cm to throw it to a height of 5m.\left(c\right) Since 0.5*k*(0.03{)}^2 = 5\Rightarrow k=11111.11$

Very crudely, centre of mass is the point where the mass appears to be concentrated. In a non-uniform gravitational field, it doesn't coincide with centre of mass. It can also lie outside the body. Since torque can be measured around any axis, it is not necessary for the centre of gravity to lie inside the body.

Centre of mass is hypothetical geometry point where whole mass of body is concentrated. It is position vector.Centre of mass may lie outside the body.

A NO FORCE BUT GRAVITATIONAL FORCE??????????

B BECAUSE THE DENSITY OF CORK IS LESS THAN WATER . FORCE =GRAVITATIONAL

(a) No force, since it is falling with constant speed.

(b) Here also the gravitational force is balanced by the upward buoyant force. Hence net force is zero.

as we know the expresssion for orbital velocity as,

$v_0= \sqrt{\frac{GM}{R+h}}where v_0= orbital velocity R= radius of sunM=mass of sunR+h= orbital radiusG= gravitational cons\tan t$

R+h= $R_0$ (Let)

now the above equation can be writtenn as in terms of circumference and time i.e., $v_0$= circumference made by earth / time taken by it.

so, $\frac{2\pi (R+h)}{t}=\sqrt{\frac{GM}{R+h}}as we already as let (R+h)=R_{0}so \frac{2\pi R_0}{t}=\sqrt{\frac{GM}{R_0}}t=\frac{2\pi R_0}{\sqrt{\frac{GM}{R_0}}}now when orbital radius is 4 times of orbital radius made by earthi.e., {R^{\text{'}}}_0 = 4R_{0}let t^{\text{'}} be the time made by unknown planetso t^{\text{'}}= \frac{2{{\mathrm{\pi R}}^{\text{'}}}_0}{\sqrt{\frac{GM}{{R^{\text{'}}}_0}}}implies that t^{\text{'}}= \frac{2\pi 4R_0}{\sqrt{\frac{GM}{4R_0}}}now \frac{t}{t^{\text{'}}}=\frac{\frac{2\pi R_0}{\sqrt{\frac{GM}{R_0}}}}{\frac{2\pi 4R_0}{\sqrt{\frac{GM}{4R_0}}}}on solving the above expression we get{t}^{\text{'}}= 8thence the required time is t^{\text{'}}= 8 X 1 years = 8 years$

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dear , if cannot explain the answer then please dont suggest any other website

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not necessary that all constant are dimensionless.

for some constant have it has dimension and some dont have

for eg:gravitational constant ,g,planck's constant etc have dimension 

adarsh pls mention height of roof it depends on height if height is 5-9 or something like that then the previous drop would have already touched the ground.

A bunch of wire is more flexible than a single one so it will carry more load

A combination of core will carry more load