A 1000 mm wide steel sheet made of C35 is normalized by cold rolling 10 mm thick
deformed to 5 mm. The rollers, 600 mm in diameter, run at a peripheral speed of 0.12 m/s.
The deformation efficiency is 55%.
Find out:
a) the roller force
b) the roller torque
c) the performance on the pair of rollers.

The correct question is;

A particular table in a relational database contains 100,000 rows, each of which requires

200 bytes of memory. Estimate the time in milliseconds to to insert a new row into the

table when each of the following indices on the related attribute is used. Assume a page

size of 4K bytes and a page access time of 20 ms.

a. No index (heap file)

b. A clustered, non-integrated B+ tree index, with no node splitting

required. Assume that each index entry occupies 100 bytes. Assume that the

index is 75% occupied and the actual data pages are 100% occupied. Assume

that all matching entries are in a single page.

Answer:

A) 20 ms

B) 120 ms

Explanation:

A) Append (at the end of file). Just one IO, i.e., 20 ms

B) Now, when we assume that each entry in the index occupies 100 bytes, then an index page can thus hold 40 entries. Due to the fact that the data file occupies 5000 pages, the leaf level of the tree must contain at least 5000/40 pages which is 125 pages.

So, the number of levels in the tree (assuming page 75% occupancy in the

index) is (log_30 (125)) + 1 = 3. Now, if we assume that the index is clustered and not integrated with the data file and all matching entries are in a single

page, then 4 I/O operations and 80ms are required to retrieve all matching

records. Two additional I/O operations are required to update the leaf page

of the index and the data page. Hence, the time to do the insertion is

120ms.

Answer:

The power produced by the turbine is 74655.936 kW.

Explanation:

A turbine is a device that operates at steady-state. Let suppose that turbine does not have heat interactions with surroundings, as well as changes in potential and kinetic energies are neglictible. Power output can be determined by First Law of Thermodynamics:

[tex]-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0[/tex]

[tex]\dot W_{out} = \dot m\cdot (h_{in}-h_{out})[/tex]

Let suppose that water enters as saturated vapor and exits as saturated liquid. Specific enthalpies are, respectively:

[tex]h_{in} = 2748.1\,\frac{kJ}{kg}[/tex]

[tex]h_{out} = 225.94\,\frac{kJ}{kg}[/tex]

The power produce by the turbine is:

[tex]\dot W_{in} = \left(29.6\,\frac{kg}{s} \right)\cdot \left(2748.1\,\frac{kJ}{kg} - 225.94\,\frac{kJ}{kg} \right)[/tex]

[tex]\dot W_{in} = 74655.936\,kW[/tex]

Answer:

V(t) = XLI₀sin(π/2 - ωt)

Explanation:

According to Maxwell's equation which is expressed as;

V(t) = dФ/dt ........(1)

Magnetic flux Ф can also be expressed as;

Ф = LI(t)

Where

L = inductance of the inductor

I = current in Ampere

We can therefore Express Maxwell equation as:

V(t) = dLI(t)/dt ....... (2)

Since the inductance is constant then voltage remains

V(t) = LdI(t)/dt

In an AC circuit, the current is time varying and it is given in the form of

I(t) = I₀sin(ωt)

Substitutes the current I(t) into equation (2)

Then the voltage across inductor will be expressed as

V(t) = Ld(I₀sin(ωt))/dt

V(t) = LI₀ωcos(ωt)

Where cos(ωt) = sin(π/2 - ωt)

Then

V(t) = ωLI₀sin(π/2 - ωt) .....(3)

Because the voltage and current are out of phase with the phase difference of π/2 or 90°

The inductive reactance XL = ωL

Substitute ωL for XL in equation (3)

Therefore, the voltage across inductor is can be expressed as;

V(t) = XLI₀sin(π/2 - ωt)

Answer:

i) VT = 52.16

VR = 147.85

ii) VT = 61

VR = 138.99

Explanation:

The step by step solution is been done, please check the attached file below to see it

Answer:

screw thrust = ML[tex]T^{-2}[/tex] 

Explanation:

thrust of a screw propeller is given by the equation = p[tex]V^{2}[/tex][tex]D^{2}[/tex] x [tex]\frac{ND}{V}[/tex]Re

where,

D is diameter

V is the fluid velocity

p is the fluid density

N is the angular speed of the screw in revolution per second

Re is the Reynolds number which is equal to  puD/μ

where p is the fluid density

u is the fluid velocity, and

μ is the fluid viscosity = kg/m.s = M[tex]L^{-1}[/tex][tex]T^{-1}[/tex]

Reynolds number is dimensionless so it cancels out

The dimensions of the variables are shown below in MLT

diameter is m = L

speed is in m/s = L[tex]T^{-1}[/tex]

fluid density is in kg/[tex]m^{3}[/tex] = M[tex]L^{-3}[/tex]

N is in rad/s = L[tex]L^{-1}[/tex][tex]T^{-1}[/tex] =

If we substitute these dimensions in their respective places in the equation, we get

thrust = M[tex]L^{-3}[/tex][tex](LT^{-1}) ^{2}[/tex][tex]L^{2}[/tex][tex]\frac{T^{-1} L}{LT^{-1} }[/tex]

= M[tex]L^{-3}[/tex][tex]L^{2}[/tex][tex]T^{-2}[/tex]

screw thrust = ML[tex]T^{-2}[/tex] 

This is the dimension for a force which indicates that thrust is a type of force

Answer:

The uniform pressure for the necessary axial force  is  W = 945 N

The uniform wear for the necessary axial force is  W = 970.15 N

Explanation:

Solution

Given that:

r₁ = 0.1 m

r₂ = 0.05m

μ = 0.35

p = 12 N or kW

N = 1500 rpm

W = 1000 N

The angular velocity is denoted as  ω= 2πN/60

Here,

ω = 2π *1500/60 = 157.07 rad/s

Now, the power transferred becomes

P = Tω this is the equation (1)

Thus

12kW = T * 157.07 rad/s

T = 76.4 N.m

Now, when we look at the uniform condition, we have what is called the torque that is frictional which acts at the frictional surface of the clutch dented as :

T = nμW R this is the equation (2)

The frictional surface of the mean radius is denoted by

R =2/3 [(r₁)³ - (r₂)³/(r₁)² - (r₂)²]

=[(0.1)³ - (0.05)³/[(0.1)² - (0.05)²]

R is =0.077 m

Now, we replace this values and put them into the equation (2)

It gives us this, 76. 4 N.m = n * 0.35* 1000 N * 0.077 m

n = 2.809 = 3

The number of pair surfaces is = 3

Secondly, we determine the uniform wear.

So, the mean radius is denoted as follows:

R = r₁ + r₂/ 2

=0.1 + 0.05/2

=0.075 m

Now, we replace the values and put it into the equation (2) formula

76. 4 N.m = n *0.35* 1000 N * 0.075 m

n= 2.91 = 3

Again, the number of pair surfaces = 3

However, for the uniform pressure with regards to the number of clutch plates is 3 we can derive the necessary axial force from the equation (2)

76. 4 N.m = 3 * 0.35 * W *0.077 m

W = 945 N

Also, for the uniform wear with regards to the number of clutch plates is 3 we can derive the necessary axial force from the equation (2)

76. 4 N.m = 3 * 0.35 * W *0.075 m

W = 970. 15 N

Answer:

The correct answer is (I) 290.81 W (II) 83.413 W (III) It is not correct to say that the rate of heat transfer must be the same to the heat of change in kinetic energy

Explanation:

Solution

Recall that:

Pressure (p) = 1.7 bar

Temperature (T₁) = 400°C which is = 673k

The velocity (v) = 50.0 m/s.

The pipe diameter (D)= 8.2 cm approximately  8.2 * 10 ^⁻2 m

The molecular air weight  (M)= 29 g/mol

Suppose Air is seen as an ideal gas

where pv = mrT

p =(m/v) r T = p = ρrT

So,

r = the characteristics of gas constant (R/m)

p = pressure

R =The universal gas constant

T = temperature

ρ = density which is (kg/m³)

R  is 8.314 J/mole -k

Then

1.7 * 10 ^5 = ρ * (8.314 /29) * 673

The density ρ = 881.09 g/ m³

(I) The mass flow rate = ρAV

thus,

m = 881.09 *π/4 ( 8.2 * 10^⁻2)² * 50

Therefore m = 232.65 g/s

We already know that

k= 1/2  mv²

k =1/2 *232.65/1000 * (50)²

so at 400°C k = 290.81 W

(II) Now in solving the  process of the constant pressure we recall that

P = ρrT

Air is cooled  to 250°C

p/r = ρT this is constant

So,

ρ₁T₂ = ρ₂T₂

881.09 * 673 = ρ₂ * 523

ρ₂ = 1133.79 g/m³

Thus,

m = ρ₂AV = 1133.79 * π /4 (8.2 * 10^ ⁻2) * 50

Hence m = 299.37 g/s

now,

k =1.2 mv² = 1.2 *(299.37)/1000 * (50)²

At 250°C, k = 374.22 W

Thus,

Δk = k ( 250°c) - k ( 400°c)

Δk = 374.22 - 290.81

Therefore,

Δk=83.413 W

(III) The steady state formula is given below

Q = W + ΔkE +ΔPE + ΔH

Now,

W = work (shaft)

Q =The rate of transfer of heat

ΔkE = The change in kinetic energy

ΔPE= The change in potential energy

ΔH =Change in enthalphy

For no shaft work, W =0

The horizontal pipe ΔPE = 0

Therefore,

The rate of heat transfer is explained as follows:

Q =ΔkE + ΔH

Because of the enthalphy,  Q is not equal to ΔkE

Finally, it is not correct to say that the rate of heat transfer must be the same to the heat of change in kinetic energy.

Answer:

Gear Y would turn Counter-Clockwise do to the opposite force created from gear X.

                          Hope this helped!  Have a great day!

I'd say number 4, number 3 looks like an exhaust valve

Answer: downward velocity = 6.9×10^-4 cm/s

Explanation: Given that the

Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m

Where radius r = 2.5 × 10^-5 m

Density = 1200 kg/m^3

Area of a sphere = 4πr^2

A = 4 × π× (2.5 × 10^-5)^2

A = 7.8 × 10^-9 m^2

Volume V = 4/3πr^3

V = 4/3 × π × (2.5 × 10^-5)^3

V = 6.5 × 10^-14 m^3

Since density = mass/ volume

Make mass the subject of formula

Mass = density × volume

Mass = 1200 × 6.5 × 10^-14

Mass M = 7.9 × 10^-11 kg

Using the formula

V = sqrt( 2Mg/ pCA)

Where

g = 9.81 m/s^2

M = mass = 7.9 × 10^-11 kg

p = density = 1200 kg/m3

C = drag coefficient = 24

A = area = 7.8 × 10^-9m^2

V = terminal velocity

Substitute all the parameters into the formula

V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]

V = sqrt[ 1.54 × 10^-9/2.25×10-4]

V = 6.9×10^-6 m/s

V = 6.9 × 10^-4 cm/s