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8090 [49]
3 years ago
13

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Engineering
2 answers:
Ainat [17]3 years ago
8 0

Answer:

meepus weepus

Explanation:

Wittaler [7]3 years ago
4 0

Answer:

I know dcdggdufug6c6 Is th best answers

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The 150-lb man sits in the center of the boat, which has a uniform width and a weight per linear foot of 3 lb>ft. Determine t
irina1246 [14]

Answer:

M = 281.25 lb*ft

Explanation:

Given

W<em>man</em> = 150 lb

Weight per linear foot of the boat: q = 3 lb/ft

L = 15.00 m

M<em>max</em> = ?

Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):

∑ Fy = 0  (+↑)     ⇒    q'*L - W - q*L = 0

⇒       q' = (W + q*L) / L

⇒       q' = (150 lb + 3 lb/ft*15 ft) / 15 ft

⇒       q' = 13 lb/ft   (+↑)

The free body diagram of the boat is shown in the pic.

Then, we apply the following equation

q(x) = (13 - 3) = 10   (+↑)

V(x) = ∫q(x) dx = ∫10 dx = 10x   (0 ≤ x ≤ 7.5)

M(x) = ∫10x dx = 5x²  (0 ≤ x ≤ 7.5)

The maximum internal bending moment occurs when x = 7.5 ft

then

M(7.5) = 5(7.5)² = 281.25 lb*ft

8 0
4 years ago
Which section of business plan should be the bulk of the plan
NeTakaya

Answer: Service and/or product line

5 0
4 years ago
QUESTION 6
Aloiza [94]
It would be 2 Portfolio
3 0
3 years ago
What does the DHCP server configures for each host?
lidiya [134]

Answer: IP address

Explanation:

DHCP server automatically assigns an IP address and other information to each host on the network so they can communicate efficiently with other endpoints

8 0
4 years ago
The rate of flow through an ideal clarifier is 8000m3 /d, the detention time is 1h and the depth is 3m. If a full-length movable
Fittoniya [83]

Answer:

a) 35%

b) yes it can be improved by moving the tray near the top

   Tray should be located ( 1 to 2 meters below surface )

   max removal efficiency ≈ 70%

c) The maximum removal will drop as the particle settling velocity = 0.5 m/h

Explanation:

Given data:

flow rate = 8000 m^3/d

Detention time = 1h

depth = 3m

Full length movable horizontal tray :  1m below surface

<u>a) Determine percent removal of particles having a settling velocity of 1m/h</u>

velocity of critical sized particle to be removed = Depth / Detention time

= 3 / 1 = 3m/h

The percent removal of particles having a settling velocity of 1m/h ≈ 35%

<u>b) Determine if  the removal efficiency of the clarifier can be improved by moving the tray, the location of the tray  and the maximum removal efficiency</u>

The tray should be located near the top of the tray ( i.e. 1 to 2 meters below surface ) because here the removal efficiency above the tray will be 100% but since the tank is quite small hence the

Total Maximum removal efficiency

=  percent removal_{above} + percent removal_{below}

= ( d_{a},v_{p} ) . \frac{d_{a} }{depth}  + ( d_{a},v_{p} ) . \frac{depth - d_{a} }{depth}  = 100

hence max removal efficiency ≈ 70%

<u>c) what is the effect of moving the tray would be if the particle settling velocity were equal to 0.5m/h?</u>

The maximum removal will drop as the particle settling velocity = 0.5 m/h

7 0
3 years ago
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