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Paha777 [63]
3 years ago
12

The line touching the circle at a point ....................... is known as ........................... .

Engineering
1 answer:
Mashutka [201]3 years ago
8 0
The line touching the circle at a point is known as the radius.
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On aircraft equipped with fuel pumps, when is the auxiliary electric driven pump used?.
pochemuha
In an airplane equipped with fuel pumps, the auxiliary electric fuel pump is used in the event the engine-driven fuel pump fails.. hope this helped !
6 0
2 years ago
4. A banking system provides users with several services:
VLD [36.1K]

A diagram showing a use case diagrams for these requirements is given in the image attached.

<h3>What is system Case diagram?</h3>

A use case diagram is known to be a kind of graphical illustration of a users in terms of their various possible association or interactions within any given system.

A use case diagram in banking can be used to prepare, depict and also to know all the functional requirements of the banking system.

Therefore, Give the use case specification for the banking system services and paying a bill online is given in the image attached.

Learn  more about Case diagram from

brainly.com/question/12975184

#SPJ1

4 0
2 years ago
A smooth concrete pipe (1.5-ft diameter) carries water from a reservoir to an industrial treatment plant 1 mile away and dischar
Kamila [148]

ANSWER:

Q = 0.17ft3/s

EXPLANATION: since the water runs downhill on a 1:100 slope, that means the flow is laminar.

Using poiseuille equation:

Q = (π × D^4 × ∆P) ÷ (128 × U × ∆X)

Q is the volume flow rate.

π is pie constant value at 3.142

D is the diameter of the pipe

∆P is the pressure drop

U is the viscosity

∆X is the length of the pipe or distance of flow.

Form the question, we are to determine U then Find Q

Therefore;

D = 1.5ft

∆P = 1pa since the minor losses are negligible.

∆X = 1mile = 5280ft.

STEP1: FIND U

Viscosity is a function of the temperature of the liquid. An increase in temperature increases the viscosity of the liquid.

We know that at room temperature, which is 25°C the viscosity of water is 8.9×10^-4pa.s . We can find the viscosity of water at 4°C by cross multiplying.

Therefore;

25°C = 8.9×10^-4pa.s

4°C = U

Cross multiply

U25°C = 4°C × 8.9×10^-4pa.s

U25°C = 0.00356°C.pa.s

Therefore;

U = 0.00356°C.pa.s ÷ 25°C

U = 1.424×10^-4pa.s

Therefore at 4°C the viscosity of water in the pipe is 1.424×10^-4pa.s

STEP2: FIND Q

Imputing the values into poiseuille equation above.

Q = (3.142 × (1.5ft)^4 × 1pa) ÷ (128 × 1.424×10^-4pa.s × 5280ft)

Q = 15.906375pa.ft4 ÷ 96.239616pa.s.ft

Therefore;

Q = 0.16547887ft3/s

Approximately;

Q = 0.17ft3/s

6 0
3 years ago
Best use for a suspension bridge
babunello [35]

Answer:

over a rive or fast moving water or canyon

Explanation: you would use a suspension bridge in an area where you can't put supports down.

4 0
3 years ago
(a)Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in leng
igor_vitrenko [27]

Answer:

a) \sigma = 12.2 (Ω-m)^{-1}

b) Resistance = 121.4 Ω

Explanation:

given data:

diameter is 7.0 mm

length 57 mm

current I = 0.25 A

voltage v = 24 v

distance between the probes is 45 mm

electrical conductivity is given as

\sigma = \frac{I l}{V \pi r^2}

\sigma  = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}

\sigma = 12.2(Ω-m)^{-1}[/tex]

b)

Resistance = \frac{l}{\sigma A}

                  = \frac{l}{ \sigma \pi r^2}

= \frac{57  \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}

Resistance = 121.4 Ω

8 0
3 years ago
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