All categories

2 years ago

Light duty scaffolds are to be used for loads up to 25 lbs./sq. ft. True False

Engineering

1 answer:

11Alexandr11 [23.1K]2 years ago

6 0

Answer:

The answer is true

Explanation:

Light duty scaffolds are to be used for loads up to 25 lbs./sq. ft.

You might be interested in

Sandwich materials typically use a high density core with non-structural cover plates. a)True b)- False

Lorico [155]

Answer: False

Explanation: Sandwich materials are usually in composite material form which has a fabrication of two thin layers which are stiff in nature and have light weighing and thick core .The construction is based on the ratio that is of stiffness to the weight .Therefore, the density of the material in the core is not high and are only connected with the skin layer through adhesive .So the given statement is false that sandwich materials typically use a high density core with non- structural cover plates.

4 0

3 years ago

Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping stati

snow_tiger [21]

Answer:

$D=0.41m$

Explanation:

From the question we are told that:

Discharge rate $V_r=0.35 m3/s$

Distance $d=4km$

Elevation of the pumping station $h_p= 140 m$

Elevation of the Exit point $h_e= 150 m$

Generally the Steady Flow Energy Equation SFEE is mathematically given by

$h_p=h_e+h$

With

$P_1-P_2$

And

$V_1=V-2$

Therefore

$h=140-150$

$h=10$

Generally h is give as

$h=\frac{0.5LV^2}{2gD}$

$h=\frac{8Q^2fL}{\pi^2 gD^5}$

Therefore

$10=\frac{8Q^2fL}{\pi^2 gD^5}$

$D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}$

$D=0.41m$

8 0

3 years ago

Why is logging done during drilling?

Solnce55 [7]

Answer:

Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible

8 0

2 years ago

Express (118)10 and (-49)10 in 8-bit binary one’s complement form and then add the numbers. What would be the representation (-0

stealth61 [152]

Answer:

$118_{10}= 0110111$ 2) $-49_{10}=110001_{2}$ 3) $0_{10}=0:16 \Rightarrow 0_{10}=0_{16}$

Explanation:

1) Expressing the Division as the summation of the quotient and the remainder

for

118, knowing it is originally a decimal form:

118:2=59 +(0), 59/2 =29 + 1, 29/2=14+1, 14/2=7+0, 7/2=3+1, 3/2=1+1, 1/2=0+1

$118_{10}= 0110111$

2) $-49_{10}$

Similarly, we'll start the process with the absolute value of -49 since we want the positive value of it. Then let's start the successive divisions till zero.

|-49|=49

49:2=24+1, 24:2=12+0,12:2=6+0,6:2=3+0,3:2=1+1,1:2=0+1

100011

$-49_{10}=110001_{2}$

3) $(-0)_{10}$