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2 years ago

A market structure in which ther a single buyer

Engineering

1 answer:

faltersainse [42]2 years ago

8 0

Answer:

Monopsony

Explanation:

Kinda like a monopoly, except instead of one producer it is one buyer.

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A 400 kg machine is placed at the mid-span of a 3.2-m simply supported steel (E = 200 x 10^9 N/m^2) beam. The machine is observe

Alenkinab [10]

Answer:

moment of inertia = 4.662 * 10^6 $mm^4$

Explanation:

Given data :

Mass of machine = 400 kg = 400 * 9.81 = 3924 N

length of span = 3.2 m

E = 200 * 10^9 N/m^2

frequency = 9.3 Hz

Wm ( angular frequency ) = 2 $\pi f$ = 58.434 rad/secs

also Wm = $\sqrt{\frac{g}{t} }$ ------- EQUATION 1

g = 9.81

deflection of simply supported beam

t = $\frac{wl^3}{48EI}$

insert the value of t into equation 1

W $m^2$ = $\frac{g*48*E*I}{WL^3}$ make I the subject of the equation

I ( Moment of inertia about the neutral axis ) = $\frac{WL^3* Wn^2}{48*g*E}$

I = $\frac{3924*3.2^3*58.434^2}{48*9.81*200*10^9}$ = 4.662 * 10^6 $mm^4$

6 0

3 years ago

To convert from the U.S. Customary (FPS) system of units to the SI system of units. A first-year engineering student records thr

LenaWriter [7]

This question is incomplete, the complete question is;

To convert from the U.S. Customary (FPS) system of units to the SI system of units. A first-year engineering student records three separate measurements as 653 lb, 69.0 mi/h, and 293 × 10⁶ ft². Suppose this engineering student has to turn in the results, but the professor only accepts results given in SI units.

Required:

What is the area measurement, 293 × 10⁶ ft², in SI units?

293 × 10⁶ ft² = ?km²

Answer:

the area measurement is 27.221 km²

Explanation:

Given the data in the question;

What is the area measurement, 293 × 10⁶ ft², in SI units

we are to the result of the measured area from ft² to km²

we know that;

1 meter = 3.2808 ft

1 km = 1000 m

1 ft = (1 / 3.2808)m

1 m = ( 1/1000 ) km

since our measured are is 293 × 10⁶ ft²

hence

A = 293 × 10⁶ × [ (1 / 3.2808)m ]²

A = 27221252.74 m²

A = 27221252.74 × [ ( 1/1000 ) km ]²

A = 27.221 km²

Therefore, the area measurement is 27.221 km²

7 0

3 years ago

Is a water proof material used around tubs and

Leviafan [203]

I think it’s hard board

4 0

3 years ago

Read 2 more answers

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

4 years ago

What are the potential hazards relating to materials handling injuries?

const2013 [10]

all 4 apply all these are saftey hazards and must be taken as such anything in a work enviorment should be taken seriously and such as hazard

5 0

3 years ago