All categories

2 years ago

The component of a fluid system where a fluid is stored, but not under pressure, is called a container.

Engineering

1 answer:

forsale [732]2 years ago

6 0

Answer:

The answer is true

please friend me thank you bye.

Explanation:

You might be interested in

-1 1/6 divided by 2 1/3

pantera1 [17]

Answer:

$-\frac{1}{2}$

Explanation:

$-1\frac{1}{6}:2\frac{1}{3}=\\ \\=-\frac{7}{6}:\frac{7}{3}\\ \\=-\frac{7}{6}*\frac{3}{7}\\ \\=-\frac{1}{2}$

4 0

3 years ago

The domain of discourse is the members of a chess club. The predicate B(x, y) means that person x has beaten person y at some po

satela [25.4K]

Answer:A. No one has ever beat Nancy.

Explanation:

The dormain of discourse in a simple language is the set of entities upon which our discussions are based when discussing about something.

The dormain of discourse is also known simply as universe, can also be said to be a set of entities o

upon which certain variables of interest in some formal treatment may range.

The dormain of discourse is generally attributed to Augustus De Morgan, it was also extensively used by George Boole in his Laws of Thought.

THE LOGICAL UNDERSTANDING OF THE THE QUESTION IS THAT NO ONE HAS EVER BEAT NANCY.

8 0

2 years ago

10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de

Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span 20.83\E ft

at mid span = 1.23\E ft

shear stress 7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

$\delta = \frac{wa^2b^2}{3EI}$

$= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}$

$=\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}$

= 20.83\E ft

at mid span

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

$= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}$

$\delta = 1.23\E ft$

shear stress

$\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi$

6 0

3 years ago

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

Wman = 150 lb

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

L = 15.00 m

Mmax = ?

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

3 years ago

What is the braks mean effictive pressure?

OverLord2011 [107]

Engine cylinder pressure

Explanation:

Brake mean effective pressure is a method to calculate the engine cylinder pressure which would give the measured brake horsepower. Brake mean effective pressure is used to identify engine efficiency regardless of capacity or engine speed.
It is used to identify engine efficiency.It is measured by means of transducers or pressure gauges.
It is the measure of engine capacity to do work.

7 0

2 years ago