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# Tutor profile: Apoorva K.

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Apoorva K.
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## Questions

### Subject:Physics (Thermodynamics)

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Question:

Compute the internal energy change and temperature change for 1 mole of an ideal monatomic gas when 1500 J of heat are added to the gas and the gas does no work and no work is done on the gas

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Apoorva K.

From, first law of thermodynamics Q=U+W Eq. 1 Q= Heat Added of Removed U=Change in Internal Energy W=Work Input or output Since, no work is done W=0 Therefore, from Eq. 1 Q=U U=1500 J For ideal gas, U=Cv*(Temperature Change) For, monoatomic gas, Cv=3/2R where, R= Gas Constant = 8.314 J/kg-K Substituting, 1500 = 3/2*8.314*Temperature Change Temperature Change= 120 K

### Subject:Physics (Fluid Mechanics)

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Question:

A large reservoir has straight pipe connected to it. Water is discharged from the reservoir through this straight pipe which has 3 in. diameter and has a length of 1200 ft. The flow rate of water is 12 cfm. The discharge end is open to the atmosphere. If the open end is 40 ft below the surface level in the reservoir, what is the Darcy friction factor? Losses other than pipe friction may be ignored.

Inactive
Apoorva K.

Applying the flow equation between the reservoir free surface and exit of the tube Z1 + (p1 / γ) + {(v1^2) / (2g)} = Z2 + (p2 / γ) + {(v2^2) / 2g)} + Hf Both ends of the system are open to the atmosphere, then p1 = p2 Ignoring losses other than pipe friction, and using the Darcy equation Hf = hf = 4f (L/D) (v^2 / 2g) where v is the velocity in the pipe. It follows that v2 = v, and since Z1 – Z2 = H (height of open end below reservoir), and v1 = 0 (free surface has zero velocity), the flow equation assumes the form H = (v^2 / 2g) + 4f(L/D) (v^2 / 2g) H = (v^2 / 2g) [1 + 4f(L/D)] Q = (12 / 60) = 0.2 cfs A = (D^2π / 4) = [{(3/12)2π} / 4] = (π / 64) v = (Q/A) = {(0.2) / (π/64)} = {(12.8) / π} fps Substituting 40 = [{(12.8)2} / {(64.4) (π2)}] [1 + 4f {(1200) / (3/12)}] from which f = 0.008

### Subject:Algebra

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Question:

Solve the equation |-2x + 2| -3 = -3

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Apoorva K.

The function |x| is defined as follows |x| = x for x>=0 = -x for x<0 Case 1 (-2x+2) >= 0 If -2x+2>=0 i.e. 2x=<2 i.e. x=<1 (-2x+2) > 0 for x=<1 Eq. 1 From definition of |x|, for |-2x+2| = -2x+2 for ((-2x+2)>=0). Substituting in equation above -2x+2-3=-3 -2x=-2 x=1 Eq. 2 From Eq. 1 and Eq. 2, x=1 is one solution. Case 2 (-2x+2) < 0 Similiarly, (-2x+2)<0 for x>1 Eq. 3 For, x>1 , |-2x+2| = -(2x+2) Substituting in equation above -(-2x+2)-3=-3 x=1. Eq. 4 From Eq. 3 and Eq. 4, there is no solution for this case. Therefore, x=1 is the only solution.

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