ENGINEERING
COLLEGE

Answer:

Answer: (d)Rotational speed of the shaft

Explanation: Shaft design is the design of a shaft that is used for defining the stresses at certain critical part of shaft. The shaft design has shaft diameter as a major part and this is determined by several factors like type of bearing , deflection, torque,safety factor etc.

But the least important factor for determining of the diameter is the rotational speed because it defines the rotation of an object around a particular axis, where is it states about the number of turns divisible by time. Therefore option(d) is the correct option.

COLLEGE

A test bottle containing just seeded dilution water has its DO level drop by 0.8 mg/L in a 5-day test. A 300 mL BOD bottle filled with 30 mL of wastewater and the rest with seeded dilution water experiences a drop of 7.3 mg/L in the same period (5-day). Calculate the BOD5 of the wastewater.

Answer:

BOD_5 = =65.8 mg/l

Explanation:

dilution water DO level = 0.8 m/l

BOD level drop to 7.3 mg/l

we know that BOD at 5th day can be clculated by using following relation

- DO drop in BOD bottle

- dilution water drop

P= 30/300 = 0.1

COLLEGE

Express (118)10 and (-49)10 in 8-bit binary one’s complement form and then add the numbers. What would be the representation (-0)10 in 16-bit binary one’s complement? (be sure to show your work).

Answer:

2) 3)

Explanation:

1) Expressing the Division as the summation of the quotient and the remainder

for

118, knowing it is originally a decimal form:

118:2=59 +(0), 59/2 =29 + 1, 29/2=14+1, 14/2=7+0, 7/2=3+1, 3/2=1+1, 1/2=0+1

2)

Similarly, we'll start the process with the absolute value of -49 since we want the positive value of it. Then let's start the successive divisions till zero.

|-49|=49

49:2=24+1, 24:2=12+0,12:2=6+0,6:2=3+0,3:2=1+1,1:2=0+1

100011

3)

The first step on that is dividing by 16, and then dividing their quotient again by 16, so on and adding their remainders. Simply put:

COLLEGE

Compute the total elongation of a steel bar, originally 25 in. long, if the induced tensile stress is 15,000 psi.

Answer:

total elongation = 0.0125 in

Explanation:

given data

length = 25 in

tensile stress = 15,000 psi

to find out

total elongation of a steel bar

solution

we take here elasticity E = 30 × psi

so we get here total elongation of steel bar that is express as

total elongation = ..........................1

put here value and we will get

total elongation =

total elongation = 0.0125 in

COLLEGE

the car travels around the circular track having a radius of r=300 m such that when it is at point A it has a velocity of 5m/s which is increasing at the rate of v =(0.06t) m/s^2, where t is in seconds. Determine the magnitudes of its velocity and accerlation when it has traveled one third of the way around the track

The magnitude of the velocity and the acceleration is 0.06 m/s^2., the velocity is 11 m/s

How to solve for the velocity and the accelerationGiven the information, we can calculate the velocity and acceleration of the car as it moves one-third of the way around the circular track.

First, we need to find the time t when the car has traveled one-third of the way around the track. The distance traveled by the car in one-third of the way around the track is d = r/3, where r is the radius of the track. The velocity of the car at time t is v = 5 + 0.06t. The distance traveled by the car can be found using the equation d = vt. Setting these two equations equal to each other, we have:

d = r/3 = 5t + 0.03t^2

Solving for t, we find that t = 20 seconds.

Next, we can find the velocity of the car at time t = 20 seconds using the equation v = 5 + 0.06t = 5 + 0.06 * 20 = 11 m/s.

Finally, the acceleration of the car can be found using the equation a = dv/dt = 0.06 m/s^2.

So, when the car has traveled one-third of the way around the track, its velocity is 11 m/s and its acceleration is 0.06 m/s^2.

Read more on velocity and acceleration here:brainly.com/question/460763

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COLLEGE

Claim: Most adults would erase all of their personal information online if they could. A software firm survey of 674674 randomly selected adults showed that 6060% of them would erase all of their personal information online if they could. Find the value of the test statistic. The value of the test statistic is

Answer:

5.192

Explanation:

This is a two tailed test hence the null and alternative hypothesis will be

0.5" alt="H_a: P>0.5" align="absmiddle" class="latex-formula">

Sample size, n=674

Sample proportion

Z test statistic formula for one sample is given by

Substituting the values given then

COLLEGE

Consider a large plane wall of thickness L = 0.4 m, thermal conductivity k = 2.3 W/m·K, and surface area A = 30 m2. The left side of the wall is maintained at a constant temperature of T1 = 90°C, while the right side loses heat by convection to the surrounding air at T[infinity] = 25°C with a heat transfer coefficient of h = 24 W/m2·K. Assuming constant thermal conductivity and no heat generation in the wall, (a) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, (b) obtain a relation for the variation of temperature in the wall by solving the differential equation, and (c) evaluate the rate of heat transfer through the wall. Answer: (c) 9045 W

Answer:

a. -k {(dT(L)/dx} = h{T(L) - T(s)}

b. T(x) = 90 - 131.09(x)

c. 9045W

Explanation:

Given detail

Thickness of wall L = 0.4 m

Left side temperature of the wall T1 = 90°C,

Thermal conductivity k = 2.3 W/m.k

Right side temperature of the wall T(infinity) = T(s) = 25°C

Heat transfer coefficient h = 24W/m2.K

(a) Differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall,

Qwall = -kA {(dT(0)/dx}

{d²T /dx²} = 0

T(0) = 90°C

-k {(dT(L)/dx} = h{T(L) - T(s)}

(b) relation for the variation of temperature in the wall by solving the differential equation.

T(x) = xC1 + C2

T(0) = 0*C1 + C2 = 90°C

T(0) = C2 = 90°C

-kC1 = hLC1 + hC2 -T(s)h

-C1 (k+hL) = h (C2 - T(s))

- C1 = {h(C2 - T(s)) / (k + hL)}

C1 = - { 24(90 - 25) / (2.3 + 24*0.4) }

C1 = - 131.09°C

T(x) = x*(-131.09) + 90

T(x) = 90 - 131.09(x)

(c) rate of heat transfer through the wall

Qwall = - 2.3 * 30 * (-131.09)

Qwall = 9045W ans

MIDDLE SCHOOL

When air and fuel are compressed they become very what?

Flammable (This question can help you to understand)

COLLEGE

A nonconducting spherical shell of inner radius R1 and outer radius R2 contains a uniform volume charge density rho throughout the shell. Use Gauss's law to derive an equation for the magnitude of the electric field at the following radial distances rrr from the center of the sphere. Your answers should be in terms of rho, R1, R2, r, ε0, and π. (a) r < R1

(b) R1 (c) r>R2

Answer:

a) 0; b) 0; c) E= ρ(R2^3-R1^3)/(3ε0r^2)

Explanation:

To solve this problem, we should carefully analyse Gauss's law. Equation of the gauss's law says, that the flux of the electric field for the given enclosed surface is equal to the total charge, which is enclosed by this surface, divided by ε0.

Note, that the flux is a multiplication of the electric field and the area of the given enclosed surface.

From this idea, we can get the following answers without mathematical analysis:

a) when the distance is less then R1, there are no charges stored inside the area, so the flux will be equal to 0:

E*A=Σq/ε0, as q=0, then E=0 V/m;

b) At the distance R1, we have similar situation- on R1 there are no charges being stored inside the surface. As a result, we can repeat the calculations from above: E*A=Σq/ε0

c) For the case, when r>R2, we have the entire charge of the sphere being stored inside the enclosed surface. In this case, the enclosed surface is a sphere, with radius r.

For such sphere, the area of it can be calculated as: A=4πr^2

To calculate electric field, we should also analyse the total charge inside the given surface. As the charge is stored in the volume, we should multiply charge's density with the given volume. Then, the total charge can be found as: Q=ρ* 4π(R2^3-R1^3)/3. Note, that the total volume of the charged material is equal to the subtraction of the volume of the sphere with the radius R1 and radius R2.

Using Gauss's Law, we can derive the following:

E*A=Q/ε0

E*4πr^2=ρ* 4π(R2^3-R1^3)/(3ε0)

From where, E=(ρ(R2^3-R1^3))/(3ε0r^2)

I would like to add, that we can also find the filed in the area between R1 and R2 (R1

COLLEGE

A fiberglass composite is composed of a matrix of vinyl ester and reinforcing fibers of E-glass. The volume fraction of E-glass is 31%. The remainder is vinyl ester. The density of the vinyl ester is 0.79 g/cm3, and its modulus of elasticity is 4.04 GPa. The density of E-glass is 3.011 g/cm3, and its modulus of elasticity is 80 GPa. A section of composite 1.00 cm by 25.00 cm by 200.00 cm is fabricated with the E-glass fibers running longitudinal along the 200 cm direction. Assume there are no voids in the composite. Determine the modulus of elasticity of the composite in GPa.

Full Question

Calculate the modulus of elasticity of the composite

I. In the longitudinal direction of the glass fibre

2. In the perpendicular direction of glass fibre

Answer:

1. 27.5876 GPa

2. 5.725 GPa

Explanation:

Given

Dimensions of the fibreglass is given by: 1.00 cm x 25.00 cm x 200.00 cm

Volume = 1 * 25 * 200 = 5000cm³

P1 = The volume fraction of the E-glass is 31%

P2 = The remainder 69% is Vinyl Ester

Volume of Vinyl Ester= 69% * 5000cm³ = 3450cm³

I. In the longitudinal direction of the glass fibre

Where

Modulus of elasticity of glass, E1 = 80 Gpa

Modulus of elasticity of vinyl ester, E2 = 4.04 Gpa

Modulus of elasticity = E1P1 + E2P2

Modulus of elasticity of vinyl esters = 31% * 80 + 69% * 4.04

Modulus = 24.8 + 2.7876

Modulus = 27.5876 GPa

ii. In the perpendicular direction of glass fibre

This is solved by

E1.E2/(P1.E2 + P2.E1)

= (80 * 4.04)/(31% * 4.04 + 69% * 80)

= 323.2/(1.2524 + 55.2)

= 5.725177317527687

= 5.725 GPa