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tiny-mole [99]
3 years ago
6

How to programe a disk

Engineering
1 answer:
Colt1911 [192]3 years ago
6 0

Answer:

insert the disk in the laptop, wait of a minute or 2 and then a folder will open in my PC.

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The upper joint is directly above the lower joint when<br> viewed from the side
Galina-37 [17]

Answer: zero caster

Explanation:

3 0
2 years ago
A motor vehicle engine operating with a diesel engine takes in atmospherics air at a temperature of 30°C and pressure of 1 bar.
nydimaria [60]

Answer:

η=0.568

Explanation:

At inlet condition

temperature = 30 C and pressure P=1 bar

The maximum temperature = 13456 C

Compression ratio r= 12

We know that for process 1-2

\dfrac{T_2}{T_1}=r^{\gamma -1}

\dfrac{T_2}{303}=12^{1.4 -1}

T_2=818.68 K

Now for process 2-3

\dfrac{T_3}{T_2}=\dfrac{V_3}{V_2}

\dfrac{273+1345}{818.86}=\dfrac{V_3}{V_2}

\dfrac{V_3}{V_2}=1.97

So the cut off ratio ρ=1.97

Efficiency of diesel engine

\eta =1-\dfrac{\rho ^{\gamma}-1}{r^{\gamma -1}\gamma \left (\rho -1\right )}

Now put the values

\eta =1-\dfrac{1.97 ^{1.4}-1}{12^{1.4-1}\times 1.4\times \left (1.97 -1\right )}

   ⇒η=0.568

So the efficiency is 56.8%.

4 0
3 years ago
The 40-ft-long A-36 steel rails on a train track are laid with a small gap between them to allow for thermal expansion. Determin
vfiekz [6]

Answer:

Ф = 0.02838 ft

F  = 1,032 N

Explanation:

To find out gap delta,

As it is case of free thermal expansion,

First we start with, some assumptions we have to made to solve this problem.

1. Thermal Expansion Coefficient of Steel is ∝= 6.45 ×10^(-6)

2. Modulas of elasticity for A-36 steel is E= 200 GPa

3. Area of rail is assumed to be unit area.

The gape required can be given by,

Ф = ∝  × ΔT  × L  ... where Ф= Gap Delta in ft

                                          ΔT= Temperature rise in F

                                               = 90- (-20)

                                               =  110 F

Ф =  6.45 ×10^(-6) × 110 × 40

Ф =  28,380 × 10^(-6) ft

Ф = 0.02838 ft     .... total gape required for expansion of steel rails

Stress induced in rails is given by,

   σ     =  ∝  × ΔT  × E

          =  6.45 ×10^(-6)   × 110  × 200

  σ      =  1,41,900 Pa

Now, let's find axial force in rails,

Here,we have to consider  ΔT= 20 F.

As due to temperature change, axial force generated in rails can be find by,

F = A × ∝ × ΔT× E × L

F = 1 × 6.45 × 10^(-6) × 20 × 200 × 10^(-9) × 40

F = 25,800 × 40 × 10^(-3)

F = 10,32,000 × 10^(-3)

F= 1,032 N

Finally, due to temperature change, rail is subjected to axial force, axial stress.

8 0
3 years ago
Pls help me answer my module
Otrada [13]

Answer:

Hand tools based on job requirement and its importance and the classification of hand tools according to its function and its importance are discussed below in details.

Explanation:

Hand tools based on work requirement is essential because Every tool is specifically invented for a particular purpose, so picking the accurate tool will also reduce the amount of energy needed to get work done right without causing injury or harm to either the tools or the exterior being worked on.

classifying of hand tools: wrenches, screwdrivers, cutters, striking tools, hammer tool or struck, pliers, vise, clamps, snips, saws, drills, and knives.

4 0
4 years ago
Air enters a 200 mm diameter adiabatic nozzle at 195 deg C, 500 kPa and 100 m/s. It exits at 85 kPa. If the exit diameter is 158
ahrayia [7]

Answer:

v_2 = 160.23 m/s

T_2 = 475.797 k

Explanation:

given data:

Diameter =d_1 = 200mm

t_1 =195 degree

p_1 =500 kPa

v_1 = 100m/s

p_2 = 85kPa

d_2 = 158mm

from continuity equation

A_1v_1 = A_2v_2

v_2 = \frac{\frac{\pi}{4}d_1^2 v_1^2}{\frac{\pi}{4}d_2^2}

v_2 = \frac{d_2v_1}{d_2^2}

v_2 = [\frac{d_1}{d_2}]^2 v_1

      = [\frac{0.200}{0.158}]^2 \times 100

v_2 = 160.23 m/s

by energy flow equation

h_1 + \frac{v_1^2}{2} +gz_1 +q =h_2 + \frac{v_2^2}{2} +gz_2 +w

z_1 =z_2 and q =0, w =0 for nozzle

therefore we have

h_1 -h_2 =\frac{v_1^2}{2} -\frac{v_2^2}{2}

dh = \frac{1}{2} (v_1^2 -v_2^2)

but we know dh = Cp dt

hence our equation become

Cp(T_2 -T_1) = \frac{1}{2} (v_1^2 -v_2^2)

Cp (T_2 -T_1) = 7836.94

(T_2 -T_1) = \frac{7836.94}{1.005*10^3}

(T_2 -T_1) = 7.797

T_2 = 7.797 +468 = 475.797 k

8 0
3 years ago
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