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3 years ago

Who is the inventor that created the light bulb

Engineering

1 answer:

vlada-n [284]3 years ago

3 0

Answer:

Thomas Elva Edison

He invented light.

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There are four spheres of earth. These include __________, _________, ________ and _________. They are important because when th

ololo11 [35]

Answer:

"lithosphere" , "hydrosphere" , "biosphere" , "atmosphere"

4 0

3 years ago

Define volume flow rate Q of air flowing in a duct of area A with average velocity V

Shalnov [3]

Answer:

The volume flow rate of air is $Q=A\times V$

Explanation:

A random duct is shown in the below attached figure

The volume flow rate is defined as the volume of fluid that passes a section in unit amount of time

Now by definition of velocity we can see that 'v' m/s means that in 1 second the flow occupies a length of 'v' meters

From the attached figure we can see that

The volume of the prism that the flow occupies in 1 second equals

$Volume=Area\times V=A\times V$

Hence the volume flow rate is $Q=V\times A$

3 0

3 years ago

The water depths upstream and downstream of a hydraulic jump is 0.3 m and 1.2 m. Determine flow rate

Sergio [31]

This question is incomplete, the complete question is;

the water depths upstream and downstream of a hydraulic jump is 0.3 m and 1.2 m. Determine flow rate ( in m³ ) if the rectangular channel is 20 m wide.

Answer:

the flow rate is 32.549 m³/sec

Explanation:

Given that

y₁ = 0.3 m

y₂ = 1.2 m

β = 20 m

Now for Rectangular Channel, we know that;

2q²/g = y₁y₂( y₁ + y₂)

where g = 9.81 m/s²

and q = Q/β

2(Q/β)²/g = y₁y₂( y₁ + y₂)

we substitute our given values

2(Q/20)²/9.81 = 0.3 × 1.2( 0.3 + 1.2)

2(Q²/400)/9.81 = 0.36(1.5)

2(Q²/400) = 0.54 × 9.81

Q²/400 = 5.2974 / 2

Q²/400 = 2.6587

Q² = 1059.48

Q = √1059.48

Q = 32.549 m³/sec

Therefore the flow rate is 32.549 m³/sec

3 0

4 years ago

A cylindrical specimen of a brass alloy having a length of 60 mm (2.36 in.) must elongate only 10.8 mm (0.425 in.) when a tensil

Gemiola [76]

The radius of the specimen is 60 mm

<u>Explanation:</u>

Given-

Length, L = 60 mm

Elongated length, l = 10.8 mm

Load, F = 50,000 N

radius, r = ?

We are supposed to calculate the radius of a cylindrical brass specimen in order to produce an elongation of 10.8 mm when a load of 50,000 N is applied. It is necessary to compute the strain corresponding to this

elongation using Equation:

ε = Δl / l₀

ε = 10.8 / 60

ε = 0.18

We know,

σ = F / A

Where A = πr²

According to the stress-strain curve of brass alloy,

σ = 440 MPa

Thus,

$sigma = 50,000 / \pi (r)^2\\\\440 X 10^6 = \frac{50,000}{3.14 X (r)^2}\\\\r = 0.06m\\r = 60mm\\\\\\$

Therefore, the radius of the specimen is 60 mm

3 0

4 years ago

No engine is perfectly efficient because?

iren2701 [21]

B. most fuel sources create environmental change (the process of making the efficient engine will still create environmental damage)

8 0

3 years ago

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