Physics Assignment Help: 30-kg hanging object is connected by a light..

Q) 30-kg hanging object is connected by a light, inextensible cord over a light, frictionless pulley to a 5.00-kg block that is sliding on a flat table. Taking the coefficient of kinetic friction as 0.162, find the tension in the string.

Sol:

It’s given that M1=5kg, M2=30kg, Also M1 is moving with acceleration a, caused by the Tension in string: T_s, and the friction, F_r, Hence aM1= T_s - uM1g, where u= coefficient of kinetic friction.

Similarly, the mass 30kg is moving with the same above acceleration a by the forces of weight and the Tension in the string,

Hence aM2=M2g- T_s

By adding the above two equations we get,

a= (M2g-uM1g)/ (M1+M2)

Therefore the tension in the string is given by,

T_s = M1g (M2-uM1)/ (M1+M2) +uM1g

By substituting the values in the above equation, we get

T_s = [5*9.8*(30-0.162*5)/ (5+30)] +0.162*5*9.8

  = 48.8N

Read and Answer: The International Centre for Radio Astronomy Research...

1) The International Centre for Radio Astronomy Research

Sol:

The International Centre for Radio Astronomy Research is a collaborative centre placed in Perth, Western Australia. They are already achieving research excellence in both astronomical science and engineering. The International Centre for Radio Astronomy Research has grown steadily from its launch in September 2009. An equal joint venture between The University of Western Australia and Curtin University, International Centre for Radio Astronomy Research has strong support from the West Australian as well as Australian governments and is working very closely with industry and research partners in Australia and in the entire world. Wireless internet (WIFI) inventor honored Curtin University has been recognized as Dr John O’Sullivan’s ground breaking work in astronomy and wireless technologies by bestowing John with an Honorary Science Doctorate. The latest telescope puts focus on high school discoveries where a new internet telescope was launched by State Education Minister Dr Elizabeth Constable at The Western University of Australia. Also
Sky Noise and Sensitivity Studies of Array Configurations and Antenna Elements were done for the Low-Frequency Band of the SKA.

Questions:

a) What is WIFI?

b) Expand ICRAR?

c) Internet telescope was launched by whom?

d) Who invented wireless internet?

Sol:

a) Wireless internet/wireless LAN.

b) International Centre for Radio Astronomy Research.

c) Dr. Elizabeth.

d) Dr John O’Sullivan

Looking for the solutions of few mathematics questions?

a) ABC' + B C' D' + B C + C' D = B + C' D

Sol:

LHS = (ABC’+BC)+ (BC’D’+C’D)

       =B(C+C’A) +C’ (D+D’B)

       =B(C+A) +C’ (D+B)

       =AB+BC+C’D+C’B

      =AB+C’D+B (C+C’)                //C+C’=1

      =AB+C’D+B

      =B (1+A) +C’D                        //1+A=1

      =B+C’D

      =RHS

b) WY + W' Y Z' + WXZ + W' X Y' = WY + W' X Z' + X' Y Z' + X Y' Z

Sol:

RHS=wy + w’xz’ + x’yz’ + xy’z

        =WY(1+XZ)+ w’xz’(Y+Y’)+ x’yz’ (W+W’)+ xy’z(W+W’)         //Y+Y’=W+W’=1+XZ=1

        = wy + wxyz + w’xyz’ + w’xy’z’ +wx’yz’ + w’x’yz’ + wxy’z + w’xy’z

        =wy + w’xyz’ + w’x’yz’ + wxyz +wxy’z + w’xy’z + w’xy’z’ + wx’yz’

        =wy + wx’yz’ + w’yz’(x + x’) + wxz(y + y’)+w’xy’(z + z’)

        =wy(1 + x’z’) + w’yz’ + wxz + w’xy’

        = WY + W' Y Z' + WXZ + W' X Y'

        =LHS

c) A D' + A' B + C'D + B'C = (A' + B' + C' + D')(A + B + C + D') = AC’+A’B +B‘C + D’

Sol:

RHS=(A' + B' + C' + D')(A + B + C + D')

        =A’A+A’B+A’C+A’D’+B’A+B’B+B’C+B’D’+C’A+C’B+C’C+C’D’+D’A+D’B+D’C+D’D’

        =A’B+A’C+A’D’+B’A+B’C+B’D’+C’A+C’B+C’D’+D’A+D’B+D’C+D’   //AA’=0; D’D’=D’

        = AC’+A’B +B‘C + D’+ A’C (B+B’) + B’A (C+C’) + C’B (A+A’)    //A+A’=1

        = AC’ (1 + B’ + B) + A’B (1 + C’ + C) +B’C(1 + A + A’) + D’    //1+A=1

        = AC’+A’B +B‘C + D’

        =LHS

Homework Help: What is the rms speed of each helium atom...

Q)The molecular mass of helium is 4 g/mol, the Boltzmann’s constant is 1.38066 ×
10−23 J/K, the universal gas constant is 8.31451 J/K · mol, and Avogadro’s number
is 6.02214 × 1023 1/mol. Given: 1 atm = 101300 Pa.
What is the rms speed of each helium atom?
Answer in units of m/s.

Sol:

The average kinetic energy is given by:

KE_atom = (PV)/N

Number of atoms (N)= Avogadro's number (A) * n
where N = A*n = A*(PV/RT)

N = 6.02214 x 10^23 atoms/mole * (96235 Pa * 0.01414 m^3) / (8.31451 J/K-mol * 285.15 K)
N = 3.456 x 10^23 atoms

KE_atom = (96235*0.01414 /3.456 x 10^23 )J/atom
      = 3.94 X 10^-21 J/atom

KE_a = (1/2) m v^2
m = (Molecular Mass (Mm) / A)
m = (0.004 kg/mole / 6.02214 x 10^23 atoms/mole)
m = 6.42 x 10^-27 kg/atom
v^2 = (2 KE_a)/m
v = sqrt((2 KE_a) / m)
v = sqrt((2 * 3.94 X 10^-21 J/atom) / 6.42 x 10^-27 kg/atom)
v = 1108 m/s

Homework Help: What is the speed of the block when the spring has been compressed...

A block with a mass of 0.148kg is traveling at 8.4m/s on a horizontal,frictionless surface when it hits a spring with a spring constant of 722 N/m. What is the compression of the spring when the block has slowed to 3.9 m/s?What is the speed of the block when the spring has been compressed by 0.049m?

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Homework Help: what is the new pressure in the tank?

Gas is confined in a tank at a pressure of 10 atm and a temperature of 13◦C. If 38 percent of the gas is withdrawn and the temperature iGas is confined in a tank at a pressure of 10 atm and a temperature of 13◦C. If 38 percent of the gas is withdrawn and the temperature is raised to 70◦C, what is the new pressure in the tank? Answer in units of atm.s raised to 70◦C, what is the new pressure in the tank?
Answer in units of atm.

Sol:

Here we will make use of the formulae:

P2 = P1*(V1/V2)*(T2/T1), where V2 = 0.62(V1), T1= 13+273= 286K

T2= 70+273= 343

P2 = 10*1.6129*343/286 = 19.34atm

Homework Help: The coefficient of performance for refrigerators....

Q) The coefficient of performance for refrigerators is defined as K = qc / w. It is also known as Tcold / Thot - Tcold.

What happens to K reversible as Tc (temp cold) goes to 0k?

Sol:

K = T_cold/(T_hot-T_cold)

K reversible means 1/K

Hence 1/K= (T_hot-T_cold)/ T_cold

              = (T_hot/ T_cold)-( T_cold/ T_cold)

         =  (T_hot/ T_cold)- 1

By neglecting 1, which is a small value, we get

1/K directly proportional to 1/ T_cold

Hence as T_cold goes to 0k, K reversible tends to infinity, a very large value.

Physics Homework: Determine the voltage across the resistor..

Q) A battery has three cells connected in series, each with an internal resistance of 0.024Ω and an emf of 1.80V . This battery is connected to a 15.0Ω resistor.

Determine the voltage across the resistor.

Sol:

Since three cells are connected in series, Hence total internal resistance will be equal to, r= 3*0.024=0.072Ω

The current flowing through the circuit is given by,

I= (EMF/R) + r, where EMF=1.80V, R=15.0Ω, r= 0.072Ω

Therefore I= (1.80/15) + 0.072

           = 0.192A

The voltage across the resistor R=15.0Ω is given by,

V_R = IR

   = 0.192*15

   =2.880V

Physics Homework: How long after Car B started the race will Car B..

Q)Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance DA beyond the starting line at t=0 . The starting line is at x=0. Car A travels at a constant speed VA. Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed VB, which is greater than VA.

1) How long after Car B started the race will Car B catch up with Car A?
Express the time in terms of given quantities.

2) How far from Car B's starting line will the cars be when Car B passes Car A?
Express your answer in terms of known quantities. (You may use as well).

Sol:

1) Time taken = Distance travelled/Velocity

Distance travelled by Car A = DA

Velocity = Difference in velocity between Car B and Car A

         = VB – VA

Therefore, Time taken t= DA/(VB-VA) secs

2) Distance travelled by Car B in time t= DA/(VB-VA),

S=(Velocity of Car B) *(time taken)

 = VB*DA/(VB-VA) mts

Physics Homework: Reduce to a single resistor..

1). Reduce to a single resistor. Go step by step and indicate the series or parallel combinations being reduced.

Sol:

Step1: R3||R4 and R6||R7

R34 = 220*470/(220+470) = 150ohms

R67 = 330*680/(330+680) = 222.2ohms

Step2: R34(series)R2 and R67(series)R5

R2s = 150+150 = 300ohms

R5s = 100+222.2 = 322.2ohms 

Step3: [R2s||R5s](series)R1

R_eq = [300*322.2/(300+322.2)]+ 47

    = 202.35ohms

2). Reduce to a single resistor. Go step by step and indicate the series or parallel combinations being reduced.

Sol:

Step1: R3(series)R5

R35= 220+100 = 320ohms

Step2: R35||R4

R4p= 320*470/(320+470)

   = 190.38ohms

Step3: R4p(series)R1

R4p1= 190.38+47= 237.38ohms

Step4: R4p1||R2

R_eq = 237.38*150/(237.38+150)= 91.92ohms

3). Reduce to a single resistor. Go step by step and indicate the series or parallel combinations being reduced.

Sol:

Step1: R2(series)R6

R26 = 150+330= 480ohms

Step2: R26||R3

R3p = 480*220/(480+220) = 150.86ohms

Step3: R3p(series)R4

R3p4 = 150.86+470 = 620.86ohms

Step4: R3p4||R5

R5p = 620.86*100/(620.86+100)= 86.13ohms

Step5: R5p||R6

R56p = 86.13*330/(86.13+330)= 68.3ohms

Step6: R56p(series)R1

R_eq = 68.3+47 = 115.3ohms

4). Find VAB:

Sol:

Step1: R4||R5

R45 = 470*100/(470+100)= 82.46ohms

Step2: R45(series)R3

R3s = 82.46+220 = 302.46ohms

Step3: R3s||R2

R3s2 = 302.46*150/(302.46+150)= 100.27ohms

Step4: R3s2(series)R1

R1s = 100.27+47 = 147.27ohms

Total current, I=30/147.27 = 0.2037A

VAB = 0.2037*100.27 = 20.43V

5). Find VAB and the power supplied by the source:

Sol:

Step1: R3||R4

R34 = 220*470/(220+470)= 150ohms

Step2: R34(series)R2

R2s = 150+150 = 300ohms

Step3: R2s||R1

R_eq = 300*47/(300+47)= 40.63ohms

VAB = 50V

Power supplied by the source, P = VI = V²/ R_eq

                           

                                              = (50*50)/40.63

                                = 61.53W

Physics Homework: What current in the solenoid would be needed to produce a field of..

Q)
What current in the solenoid would be needed to produce a field of 1.5×10−3 T but in the opposite direction?

Sol:

From question 2, magnetic field of 1.6*10^-3, produces 2400 turns

Hence, magnetic field of 1.5*10^-3, produces 2250 turns

Magnetic field, B=u_o*n*I

Current, I = 1.5*10^-3/(4*pi*10^-7*2250)= 0.53A

Physics Homework: Find the current in the..

Q) 57.0Ω resistor is connected in parallel with a 123.0Ω resistor. This parallel group is connected in series with a 24.0Ω resistor. The total combination is connected across a 15.0-V battery. Find
(a) the current in the 123.0Ω resistor(in A)

(b) the power dissipated in the 123.0Ω resistor(in W).

Sol:

Equivalent Resistance, R_eq= (57||123)series(24)

                          =[57*123/(57+123)]+23 = 61.95Ω

a) Total current in the combination, I= 15/61.95

= 0.242A

Voltage across parallel combination will be,

V=0.242*38.95= 9.43V

Hence current in 123ohm resistor is, I = 9.43/123= 0.0766A

b) Power dissipated, P= 9.43*0.0766 = 0.723W

Physics Homework: What is the potential difference across the three resistors?

Q) Three resistors, 26, 48, and 76 , are connected in series, and a 0.49 A current passes through them. What is

(a) the potential difference across the three resistors(in V)?

Sol:

Since the resistances are connected in series, the equivalent resistance will be R_eq = 26+48+76 = 150ohms

Total Voltage, V= I*R_eq = 150*0.49 = 73.5V

1) Potential difference across 26 = 26*0.49= 12.74V

2) Potential difference across 48 = 48*0.49= 23.52V

3) Potential difference across 76 = 76*0.49= 37.24V

Physics Homework: find the speed of the block after it has moved?

Q) 19 kg block initially at rest pulled to the right along a horizontal surface by a constant, horizontal force of 16.1 N. the coefficient of kinetic friction is 0.145. the acceleration of gravity is 9.8 m/s^2.find the speed of the block after it has moved 3.35 m.

Sol:

As per the concept that the work done by the force is equal to the Kinetic energy of the block, we have

u*(mg)*s=1/2mv² where work done= u*(mg)*s, u=0.145,m=19kg,g=9.8m/s², s=3.35m

Therefore by substituting the values in the above equation, we get,

0.145*9.8*3.35=0.5* v²

Hence Speed of the block after it has moved 3.35m is,

V= 3.086m/s

Physics Homework: Find the acceleration of the two blocks?

Q) In the drawing, the weight of the block on the table is 432 N and that of the hanging block is 220 N. Ignore all frictional effects, and assuming the pulley to be massless.

(a) Find the acceleration of the two blocks.
m/s2

(b) Find the tension in the cord.
N

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Physics Homework: What is the magnitude of the lift force?

The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W = 59500 N. The lift force generated by the rotating blade makes an angle of 21.0° with respect to the vertical.

(a) What is the magnitude of the lift force?


(b) Determine the magnitude of the air resistance that opposes the motion.

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TTH50010

Briefly explain the term concyclic.

Hint*: Maths - Geometry

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TTH50002

Calculate the arithmetic mean of 3,5,7,9, and 11?

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