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4 years ago

1. A farmer had 752 sheep and took one shot that got them all. How did he do it?

Engineering

1 answer:

Step2247 [10]4 years ago

3 0

Answer:

In number 752 subtract 5 and 2 from 7 i.e:7-5-2=0 it becomes zero.

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Water vapor at a rate of 40,000 kg/h and at 8 MPa and 500 C enters an adiabatic turbine and leaves the turbine at 40 kPa as satu

Minchanka [31]

Answer:

M(dot)= 40000 kg/h
Pressure= 8Mpa and T= 500C
If its adiabatic then you know that it loses no heat or Q(dot) =0
You have a turbine which drives a shaft producing work for the other systems.
We want to find entropy production of alpha.

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3 years ago

5. Bar stock of initial diameter = 90 mm is drawn with a draft = 10 mm. The draw die has an entrance angle = 18, and the coeffi

yulyashka [42]

Answer:

a) 0.2099

b) 46.5 MPa

c) 233765 N

d) 3896 W

Explanation:

r = (A'' - A') / A'', where

A'' = 1/4 * π * D''²

A'' = 1/4 * 3.142 * 90²

A'' = 6362.55 mm²

D' = D'' - d = 90 - 10 = 80 mm

A' = 1/4 * π * D'²

A' = 1/4 * 3.142 * 80²

A' = 5027.2 mm²

r = (A'' - A') / A'

r = (6362.55 - 5027.2) / 6362.55

r = 1335 / 6362.55

r = 0.2099

Draw stress = σd

Y' = k = 105 MPa

Φ = 0.88 + 0.12(D/Lc), where

D = 0.5 (90 + 80) = 85 mm

Lc = 0.5 [(90 - 80)/sin 18] = 16.18 mm

Φ = 0.88 + 0.12(85/16.18) = 1.51

σd = Y' * (1 + μ/tan α) * Φ * In(A''/A')

σd = 105 * (1 + 0.08/tan18) * 1.51 * In(6362.55/5027.2)

σd = 105 * 1.246 * 1.51 * 0.2355

σd = 46.5 MPa

F = A' * σd

F = 5027.2 * 46.5

F = 233764.8 N

P = 233764.8 (1 m/min)

P = 233764.8 Nm/min

P = 3896.08 Nm/s

P = 3896.08 W

6 0

4 years ago

Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the br

vagabundo [1.1K]

Answer:

Explanation:

Using the kinematics equation $v = v_o + a_ct$ to determine the velocity of car B.

where;

$v_o =$ initial velocity

$a_c$ = constant deceleration

Assuming the constant deceleration is = -12 ft/s^2

Also, the kinematic equation that relates to the distance with the time is:

$S = d + v_ot + \dfrac{1}{2}at^2$

Then:

$v_B = 60-12t$

The distance traveled by car B in the given time (t) is expressed as:

$S_B = d + 60 t - \dfrac{1}{2}(12t^2)$

For car A, the needed time (t) to come to rest is:

$v_A = 60 - 18(t-0.75)$

Also, the distance traveled by car A in the given time (t) is expressed as:

$S_A = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2$

Relating both velocities:

$v_B = v_A$

$60-12t = 60 - 18(t-0.75)$

$60-12t =73.5 - 18t$

$60- 73.5 = - 18t+ 12t$

$-13.5 =-6t$

t = 2.25 s

At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars

i.e.

$S_B = S_A$

$d + 60 t - \dfrac{1}{2}(12t^2) = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2$

$d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60 * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2$

d + 104.625 = 114.75

d = 114.75 - 104.625

d = 10.125 ft

3 0

3 years ago

The disk of radius 0.4 m is originally rotating at ωo=4 rad/sec. If it is subjected to a constant angular acceleration of α=5 ra

Lina20 [59]

Answer:32.4m/ $s^2$

Explanation:

Given data

$radius\left ( r\right )$ =0.4m

Intial angular velocity $\left ( \omega_0\right )$ =4rad/s

angular acceleration $\left ( \alpha\right )$ =5rad/ $s^2$

angular velocity after 1 sec

$\omega$ = $\omega_0$ + $\alpha\times\t$

$\omega$ =4+5 $\left ( 1\right )$

$\omega$ =9rad/s

Velocity of point on the outer surface of disc $\left ( v\right )$ = $\omega_0\timesr$

v= $9\times0.4$ m/s=3.6m/s

Normal component of acceleration $\left ( a_c\right )$ = $\frac{v^2}{r}$

$a_c$ = $\frac{3.6\times3.6}{0.4}$ =32.4m/ $s^2$

3 0

4 years ago

Fill in the tables and find the equivalent resistance for the following circuits:

Dafna11 [192]

Answer:

12 32

Explanation:

3 0

2 years ago