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2 years ago

A(n) ______ is used to measure fluid flow in engineering

Engineering

1 answer:

Arte-miy333 [17]2 years ago

3 0

Answer:

A pitot tube is used to measure fluid flow in engineering

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Help with this pls!!!

valentinak56 [21]

Answer: Your answer is B

Explanation:

8 0

3 years ago

Show that -40 F is approximately equal to -40 C.

V125BC [204]

Answer: Please see below the explanation.

Explanation:

The Celsius (ºC) and Farenheit (ºF), are different temperature scales, but both have identifiable values for two important physical phenomena, water freezing point and water boiling point.

For º C, these are follows:

Freezing Point= 0º C Boiling Point: 100ºC

For º F, we have the following:

Freezing Point= 32º F Boiling Point: 212 ºF

Assuming that all steps (in each scale) between two following values are equal, we have 100 º C between both points in ºC, and 190º in ºF, so degrees in Farenheit are "shorter" than in Celsius, being the relationship, approximately the following:

1 º C = 9/5 ºF.

Taking into account that, by definition, 0ºC = 32º F, we can use the following equations in order to convert ºC to ºF, and viceversa:

ºC = (ºF-32) 5/9

ºF = 32 + 9/5ºC

Now, let's calculate -40ºF, in ºC:

ºC= (-40-32)* 5/9 = -40º C

Let's do the same, converting -40ºC to ºF:

ºF = 32 + (-40)*9/5 = 32 + (-72) = -40ºF.

8 0

3 years ago

MITM can present in two forms: eavesdropping and manipulation. Discuss the process involved when an attacker is eavesdropping an

Nikitich [7]

Answer / Explanation:

Eavesdropping attack is also sometimes refereed to as sniffing attack. It is simply the process by which an attacker or hacker tries to penetrate very suddenly into an unaware individuals network or server with the intention to steal information transmitted over the network or server through that computer.

To prevent such attack, there are several mean which include installing network monitoring software to check who else is connected to the network but the most common method of preventing such attack is to encrypt the Hypertext Transfer Protocol (http) and the way to do this is by securing it with a sort of security key.

On installing the security key, the network becomes encrypted and secured such that whatever network transmitted over the network becomes encrypted and unable to read. The protocol then converts to (https).

5 0

3 years ago

The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.

zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft $J\tau$

shaft $J\tau$ = $\dfrac{\pi}{2}r^4$

where ;

r = 1 in /2

r = 0.5 in

shaft $J \tau$ = $\dfrac{\pi}{2} \times 0.5^4$

shaft $J\tau$ = 0.098218

Now; the angle of twist at B with respect to D is calculated by using the expression

$\phi_{B/D} = \sum \dfrac{TL}{JG}$

$\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

where;

$T_{CD} \ \ and \ \ L_{CD}$ are the torques at segments CD and length at segments CD

${T_{BC} \ \ and \ \ L_{BC}}$ are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

$L_{BC}}$ = 2.5 in

$J\tau$ = 0.098218

G = 11 × 10⁶ lb/in²

$T_{BC$ = -60 lb.ft

$T_{CD$ = 0 lb.ft

$L_{CD$ = 5.5 in

$\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}$

$\phi_{B/D} = \dfrac{-21600}{1079980}$

$\phi_{B/D} =$ − 0.02 rad

To degree; we have

$\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{B/D} = -1.15^0}$

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For the angle of twist of C with respect to D; we have:

$\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{C/D} = \dfrac{21600}{1079980}$

$\phi_{C/D} =$ 0.02 rad

To degree; we have

$\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{C/D} = 1.15^0}$

3 0

3 years ago

A road has a crest curve, where the PVI station is a 71 35. The road transitions from a 2.1% grade to a -3.4% grade. The highest

sveticcg [70]

Answer:

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

G1 = +2.1% ; G2 = -3.4%

Highest point of curve at station = 74 + 10

General equation of a curve:

$y = ax^{2} +bx+c\\dy/dx=2ax+b\\$

At highest point of the curve $dy/dx=o$

$2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275$

$-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\$

Station PVT

$Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)$

3 0

3 years ago