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NikAS [45]
2 years ago
7

Preheat and postheating are necessary when welding gray cast iron. * True False

Engineering
1 answer:
Zepler [3.9K]2 years ago
5 0
True will be your answer have a great day
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In order to protect yourself if you have a dispute with another drivers insurance company you should:
klio [65]

Answer:

  C. Get the names and addresses of witness to the crash

Explanation:

The best approach is to let your insurance company handle the dispute. Since that is not an option here, the best thing you can do is make sure you know who the witnesses are, so your insurance company can call upon them as needed.

8 0
2 years ago
If the rotational speed of a pump motor is reduced by 35%, what is the effect on the pump performance in terms of capacity, head
FinnZ [79.3K]

Answer:

- the capacity of the pump reduces by 35%.

- the head gets reduced by 57%.

the power consumption by the pump is reduced by 72%

Explanation:

the pump capacity is related to the speed as speed is reduces by 35%

so new speed is (100 - 35) = 65% of orginal speed

speed Q ∝ N ⇒ Q1/Q2 = N1/N2

Q2 = (N2/N1)Q1    

Q2 = (65/100)Q1

which means that the capacity of the pump is also reduces by 35%.

the head in a pump is related by

H ∝ N² ⇒ H1/H2 = N1²/N2²

H2 = (N2N1)²H1

H2 = (65/100)²H1 = 0.4225H1

so the head gets reduced by 1 - 0.4225 = 0.5775 which is 57%.

Now The power requirement of a pump is related as

P ∝ N³ ⇒ P1/P2 = N1³/N2³

P2 = (N2/N1)³P1

H2 = (65/100)²P1 = 0.274P1

So the reduction in power is 1 - 0.274 = 0.725 which is 72%

Therefore for a reduction of 35% of speed there is a reduction of 72% of the power consumption by the pump.  

8 0
2 years ago
Unit for trigonometric functions is always "radian". 1. 10 points: Do NOT submit your MATLAB code for this problem (a) Given f(x
RoseWind [281]

Answer:

Below is the required code.

Explanation:

%% Newton Raphson Method

clear all;

clc;

x0=input('Initial guess:\n');

x=x0;

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=10;

i=0;

cc=input('Condition of convergence:\n');

while ep>=cc

i=i+1;

temp=x;

x=x-(f/g);

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=abs(x-temp);

fprintf('x = %6f and error = %6f at iteration = %2f \n',x,ep,i);

end

fprintf('The solution x = %6f \n',x);

%% End of MATLAB Program

Command Window:

(a) First Root:

Initial guess:

1.5

Condition of convergence:

0.01

x = -1.815662 and error = 3.315662 at iteration = 1.000000

x = -0.644115 and error = 1.171547 at iteration = 2.000000

x = 0.208270 and error = 0.852385 at iteration = 3.000000

x = 0.434602 and error = 0.226332 at iteration = 4.000000

x = 0.451631 and error = 0.017029 at iteration = 5.000000

x = 0.451732 and error = 0.000101 at iteration = 6.000000

The solution x = 0.451732

>>

Second Root:

Initial guess:

3.5

Condition of convergence:

0.01

x = 3.300299 and error = 0.199701 at iteration = 1.000000

x = 3.305650 and error = 0.005351 at iteration = 2.000000

The solution x = 3.305650

>>

(b) Guess x=0.5:

Initial guess:

0.5

Condition of convergence:

0.01

x = 0.450883 and error = 0.049117 at iteration = 1.000000

x = 0.451732 and error = 0.000849 at iteration = 2.000000

The solution x = 0.451732

>>

Guess x=1.75:

Initial guess:

1.75

Condition of convergence:

0.01

x = 227.641471 and error = 225.891471 at iteration = 1.000000

x = 218.000998 and error = 9.640473 at iteration = 2.000000

x = 215.771507 and error = 2.229491 at iteration = 3.000000

x = 217.692636 and error = 1.921130 at iteration = 4.000000

x = 216.703197 and error = 0.989439 at iteration = 5.000000

x = 216.970438 and error = 0.267241 at iteration = 6.000000

x = 216.971251 and error = 0.000813 at iteration = 7.000000

The solution x = 216.971251

>>

Guess x=3.0:

Initial guess:

3

Condition of convergence:

0.01

x = 3.309861 and error = 0.309861 at iteration = 1.000000

x = 3.305651 and error = 0.004210 at iteration = 2.000000

The solution x = 3.305651

>>

Guess x=4.7:

Initial guess:

4.7

Condition of convergence:

0.01

x = -1.916100 and error = 1.051861 at iteration = 240.000000

x = -0.748896 and error = 1.167204 at iteration = 241.000000

x = 0.162730 and error = 0.911626 at iteration = 242.000000

x = 0.428332 and error = 0.265602 at iteration = 243.000000

x = 0.451545 and error = 0.023212 at iteration = 244.000000

x = 0.451732 and error = 0.000187 at iteration = 245.000000

The solution x = 0.451732

>>

Explanation:

The two solutions are x =0.451732 and 3.305651 within the range 0 < x< 5.

The initial guess x = 1.75 fails to determine the solution as it's not in the range. So the solution turns to unstable with initial guess x = 1.75.

7 0
3 years ago
An electric motor is to be supported by four identical mounts. Each mount can be treated as a linear prevent problems due requir
Artyom0805 [142]

GIVEN:

Amplitude, A = 0.1mm

Force, F =1 N

mass of motor, m = 120 kg

operating speed, N = 720 rpm

\frac{A}{F} =  \frac{0.1\times 10^{-3}}{1} = 0.1\times 10^{-3}

Formula Used:

A = \frac{F}{\sqrt{(K_{t} - m\omega ^{2}) +(\zeta \omega ^{2})}}

Solution:

Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K

We know that:

\omega = \frac{2 \pi\times N}{60}

so,

\omega = \frac{2 \pi\times 720}{60} = 75.39 rad/s

Using the given formula:

Damping is negligible, so, \zeta = 0

\frac{A}{F} will give the tranfer function

Therefore,

\frac{A}{F} = \frac{1}{\sqrt{(4K - 120\ ^{2})}}

0.1\times 10^{-3} =  \frac{1}{\sqrt{(4K - 120\ ^{2})}}

Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm

8 0
2 years ago
A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).
Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress \sigma_m = 90MPa

stress amplitude \sigma_a = 190MPa

a) \sigma_m =\frac{\sigma_max+\sigma_min}{2}

    90 =\frac{\sigma_{max}+\sigma_{min}}{2} --------------1

\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}

   190 = \frac{\sigma_{max}-\sigma_{min}}{2} -----------2

solving 1 and 2 equation we get

\sigma_{max} = 280MPa

b) \sigma_{min} = - 100MPa

c)

stress ratio=\frac{\sigma_{min}}{\sigma_{max}}

=\frac{-100}{280} = -0.35

d)magnitude of stress range

                      =(\sigma_{max} -\sigma_{min})

                       = 280 -(-100) = 380 MPa

3 0
3 years ago
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