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3 years ago

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Fig. 4 shows a simply-supported beam with supports A and B. The beam is subjected to three forces, 2000 N, 4000 N and 1500 N in

the directions shown.
Calculate the vertical reaction forces at supports A and B.

1 answer:

Drupady [299]3 years ago

6 0

Answer:A

Explanation:

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2. BCD uses 6 bits to represent a symbol. a) True b) False

Goshia [24]

Answer:

true because BCD used 6 bits to represent a symbol .

Explanation:

mark me brainlist

4 0

3 years ago

Which system of linear inequalities is represented by the graph? y > x – 2 and y x + 1 y x + 1 y > x – 2 and y < x + 1

KatRina [158]

Answer:

The graph representing the linear inequalities is attached below.

Explanation:

The inequalities given are :

y>x-2 and y<x+1

For tables for values of x and y and get coordinates to plot for both equation.

In the first equation;

y>x-2

y=x-2

y-x = -2

The table will be :

x y

-2 -4

-1 -3

0 -2

1 -1

2 0

The coordinates to plot are : (-2,-4) , (-1,-3), (0,-2), (1,-1) ,(2,0)

Use a dotted line and shade the part right hand side of the line.

Do the same for the second inequality equation and plot then shade the part satisfying the inequality.

The graph attached shows results.

5 0

4 years ago

Read 2 more answers

Water vapor at 100 psi, 500 F and a velocity of 100 ft./sec enters a nozzle operating at steady sate and expands adiabatically t

almond37 [142]

Answer:

a)exit velocity of the steam, V2 = 2016.8 ft/s

b) the amount of entropy produced is 0.006 Btu/Ibm.R

Explanation:

Given:

P1 = 100 psi

V1 = 100 ft./sec

T1 = 500f

P2 = 40 psi

n = 95% = 0.95

a) for nozzle:

Let's apply steady gas equation.

$h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2}$

h1 and h2 = inlet and exit enthalpy respectively.

At T1 = 500f and P1 = 100 psi,

h1 = 1278.8 Btu/Ibm

s1 = 1.708 Btu/Ibm.R

At P2 = 40psi and s1 = 1.708 Btu/Ibm.R

1193.5 Btu/Ibm

Let's find the actual h2 using the formula :

$n = \frac{h_1 - h_2*}{h_1 - h_2}$

$n = \frac{1278.8 - h_2*}{1278.8 - 1193.5}$

solving for h2, we have

$h_2 = 1197.77 Btu/Ibm$

Take Btu/Ibm = 25037 ft²/s²

Using the first equation, exit velocity of the steam =

$(1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}$

Solving for V2, we have

V2 = 2016.8 ft/s

b) The amount of entropy produced in BTU/ lbm R will be calculated using :

Δs = s2 - s1

Where s1 = 1.708 Btu/Ibm.R

At h2 = 1197.77 Btu/Ibm and P2 =40 psi,

S2 = 1.714 Btu/Ibm.R

Therefore, amount of entropy produced will be:

Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R

= 0.006 Btu/Ibm.R

3 0

3 years ago

The pressure gauge on a tank of CO2 used to fill Dora-water bottles reads 51.0 psi.

tensa zangetsu [6.8K]

Known :

Pg = 51 psi

Patm = 28 inHg = 13.7 psi

Solution :

Absolute pressure is gauge pressure plus the atmospheric pressure :

Pabs = Pg + Patm

Pabs = 51 + 13.7

Pabs = 64.8 psi

3 0

3 years ago

A negative normal strain can be considered to increase or decrease volume depending on the coordinate system used. a)True b)- Fa

geniusboy [140]

Answer:

The given statement "A negative normal strain can be considered to increase or decrease volume depending on coordinate system used" is

b) False

Explanation:

Normal strain refers to the strain due to normal stress which is when the applied stress is perpendicular to the surface.

Negative normal strain results in compression or contraction further leading to a decrease in volume while a positive normal strain results in elongation thus giving rise to an increase in volume.

6 0

3 years ago