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3 years ago

Who is Robert Goldard

Engineering

2 answers:

balu736 [363]3 years ago

8 0

Answer:

An am American engineer

maks197457 [2]3 years ago

8 0

Robert Hutchings Robert Hutchings Goddard was Associate in Nursing yank engineer, professor, physicist, and artificer who is attributable with making and building the world' initial liquid-fueled rocket. Goddard with success launched his rocket on March 16, 1926, that ushered in an era of area flight and innovation

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9) A construction company employs 2 sales engineers. Engineer 1 does the work in estimating cost for 70% of jobs bid by the comp

Irina18 [472]

The probability that the error occurred when Engineer 2 made the mistake is 0.462 on the other hand the probability that the error occurred when the engineer 1 made the mistake is 0.538

Explanation:

Let $E_{1}$ denote the event that the 1st engineer does the work.so we write

$P(E)_{1}$ =0.7

Let $E_{2}$ denote the event that the 2nd engineer does the work .So we write

$P(E)_{2}$ =0.3

Let O denote the event during which the error occurred .so we write

$P(O/E_{1} )$ =0.02(GIVEN)

$P(O/E_{2} )$ =0.04(GIVEN)

The probability that the error occurred when the first engineer performed the work is $P(E_{1} /O)$

The probability that the error occurred when the first engineer performed the work is $P(E_{2} /O)$

Now we need to find when did the error in the work occur so we will compare the probability of the work done by engineer 1 and engineer 2

lets find the Probability of the Engineer 1

Using Bayes theorem,we get

$P(E_{1} /O)$ =0.02*0.7/0.02*0.7+0.04*0.3 = 0.014/0.026=0.538

lets find the Probability of the Engineer 2

$P(E_{2} /O)$ =0.04*0.3/0.02*0.7+0.04*0.3=0.012/0.026=0.462

Since ,0.462<0.538 so it is more prominent that the Engineer 1 did the work when the error occurred

3 0

3 years ago

A motor cycle is moving up an incline of 1 in 30 at a speed of 80 km/h,and then suddenly the engine shuts down.The tractive resi

Ad libitum [116K]

Explanation:

option iii is the right answer.

4 0

3 years ago

Limestone scrubbing is used to remove SO2 in a flue gas desulfurization (FGD) system. Relevant reactions are given below. A lime

lord [1]

Answer:

hello some data related to your question is missing attached below is the missing data

answer : 63700 Ib/hr

Explanation:

Given data :

Limestone mix : consists of 94% CaCO3 and 6% inert material

Actual feed rate = 36,000 Ib/hr

SO2 in flue gas = 20,314 Ib/hr

FGD efficiency = 97%

resulting sludge contains 58% solids

Calculate the Total sludge production rate

First : determine SO2 removed in sludge

= 0.97 * 20314

= 19704.58 Ib/hr

next : moles of SO2 removed

= 19704.58 / 64 Ib/ Ib mol

= 307.88 Ib mol / hr

also moles of CaSO3 produced = 307.88 Ib mol / hr

mass of CaSO3 = 307.88 * 120 = 36946.09 Ib/hr

Therefore Total sludge production rate

= 36946.09 / 0.58

= 63700.15 Ib/hr

7 0

2 years ago

Calculate the equivalent capacitance of the three series capacitors in Figure 12-1

GrogVix [38]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Calculate the equivalent capacitance of the three series capacitors in Figure 12-1

a) 0.01 μF

b) 0.58 μF

c) 0.060 μF

d) 0.8 μF

Answer:

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

Explanation:

Please refer to the attached Figure 12-1 where three capacitors are connected in series.

We are asked to find out the equivalent capacitance of this circuit.

Recall that the equivalent capacitance in series is given by

$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

Where C₁, C₂, and C₃ are the individual capacitance connected in series.

C₁ = 0.1 μF

C₂ = 0.22 μF

C₃ = 0.47 μF

So the equivalent capacitance is

$\frac{1}{C_{eq}} = \frac{1}{0.1} + \frac{1}{0.22} + \frac{1}{0.47}$

$\frac{1}{C_{eq}} = \frac{8620}{517}$

$C_{eq} = \frac{517}{8620}$

$C_{eq} = 0.0599$

Rounding off yields

$C_{eq} = 0.060 \: \mu F$

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

5 0

3 years ago

3–102 One of the common procedures in fitness programs is to determine the fat-to-muscle ratio of the body. This is based on the

gayaneshka [121]

Answer:

x_fat = [ 0.5*(Wsa + Wsw) - p_muscle*V ] / V*( p_fat - p_muscle )

Explanation:

Given:

- The total volume of body = V

- The average density of the body = p_avg

- The density of muscle = p_muscle

- The density of fat = p_fat

Find:

Obtain a relation for the volume fraction of body fat x_fat

Solution:

- The volume of the fat is given by:

V_fat = x_fat*V

- The volume of the muscle is given by:

V_muscle = V - V_fat

= V - x_fat*V

=V*( 1 - x_fat )

- We will use the conservation of mass for the body related as:

mass_fat + mass_muscle = Total average mass

p_fat*V_fat + p_muscle*V_muscle = p_avg*V

p_fat*x_fat*V + p_muscle*V*( 1 - x_fat ) = p_avg*V

p_fat*x_fat + p_muscle*( 1 - x_fat ) = p_avg

- To determine p_1 we weigh the body in air:

Weight reading (Wsa) = m = p_1*V

p_1 = Wsa / V*g

- To determine p_2 we weigh the body in water:

Weight reading (Wsw) = m - p_w*V= p_1*V - p_w*V

Weight reading (Wsw) = V*(p_1 - p_w) = V*(p_2)

Where, p_2 = p_1 - p_water

p_2 = Wsw / V

- The average density p_avg:

p_avg = 0.5*(p_1 + p_2)

p_avg = 0.5*(Wsa / V + Wsw / V)

p_avg = 0.5*(Wsa + Wsw) / V

- Plug in the mass equation:

p_fat*x_fat + p_muscle*( 1 - x_fat ) = 0.5*(Wsa + Wsw) / V

x_fat*( p_fat - p_muscle ) = 0.5*(Wsa + Wsw) / V - p_muscle

x_fat = [ 0.5*(Wsa + Wsw) - p_muscle*V ] / V*( p_fat - p_muscle )

6 0

3 years ago