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valina [46]
3 years ago
6

Which excerpt from "The Chrysanthemums' best reveals that Elisa is proud of her

Engineering
1 answer:
Alisiya [41]3 years ago
4 0
Scissors to avoid any rough edges
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Ben leads a team of a few engineers at a robotics firm. A couple of them would like to improve their skills by taking additional
Anit [1.1K]

Answer:

i dont know

Explanation:

4 0
3 years ago
Fully developed conditions are known to exist for water flowing through a 25-mm-diameter tube at 0.01 kg/s and 27 C. What is the
Irina18 [472]

Answer:

0.0406 m/s

Explanation:

Given:

Diameter of the tube, D = 25 mm = 0.025 m

cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²

Mass flow rate = 0.01 kg/s

Now,

the mass flow rate is given as:

mass flow rate = ρAV

where,

ρ is the density of the water = 1000 kg/m³

A is the area of cross-section of the pipe

V is the average velocity through the pipe

thus,

0.01 = 1000 × 4.9 × 10⁻⁴ × V

or

V = 0.0203 m/s

also,

Reynold's number, Re = \frac{VD}{\nu}

where,

ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s

thus,

Re = \frac{0.0203\times0.025}{0.833\times10^{-6}}

or

Re = 611.39 < 2000

thus,

the flow is laminar

hence,

the maximum velocity =  2 × average velocity = 2 × 0.0203 m/s

or

maximum velocity = 0.0406 m/s

5 0
3 years ago
Discuss the difference between a dual split and a diagonal split master cylinder.
Anvisha [2.4K]

Answer:

The difference between the dual split Master cylinder and diagonal split Master cylinder is dual split it makes power brakes less prone to failure while diagonal split is when the left rear and right front brakes are on one hydraulic line while the right front and left rear brakes are on another.

7 0
2 years ago
A Venturi meter (see below) uses the principles of the Bernoulli equation to measure the velocity of a flow in a reactor system.
LenaWriter [7]

Answer:

Q=0.000604 m³/s

Explanation:

Given that

d₁=5 cm

d₂=1 cm

P= 30 KPa

Density of water ,ρ=1000 kg/m³

As we know that volume flow rate Q given as

Q=A_1A_2\sqrt{\dfrac{\dfrac{2\Delta P}{\rho}}{A_1^2-A_2^2}}

A_1=\dfrac{\pi}{4}\times 0.05^2\ m^2

A₁=0.0019 m²

A_2\dfrac{\pi}{4}\times 0.01^2\ m^2

A₂=0.000078 m²

Q=0.0019 \times 0.000078 \sqrt{\dfrac{\dfrac{2\times 30\times 1000}{1000}}{0.0019^2-0.000078^2}}\ m^3/s

Q=0.000604 m³/s

7 0
3 years ago
Disk A has a mass of 8 kg and an initial angular velocity of 360 rpm clockwise; disk B has a mass of 3.5 kg and is initially at
garri49 [273]

Answer:

A. αa= 9.375 rad/s^2, αb= 28.57 rad/s^2

B. ωa=251rpm, ωb=333rpm

Explanation:

Mass of disk A, m1= 8kg

Mass of disk B, m2= 3.5kg

The initial angular velocity of disk A= 360rpm

The horizontal force applied = 20N

The coefficient of friction μk = 0.15

While slipping occurs, a frictional force is applied to disk A and disk B

T= 1/2Mar^2a

T=1/2 (8kg) (0.08)^2

T= 0.0256kg-m^2

N=P=20N

F= μN

F= 0.15 × 20

F=3N

We have

Summation Ma= Summation(Ma) eff

Fra= Iaαa

(3N)(0.08m)= (0.0256kg-m^2)α

αa= 9.375 rad/s^2

The angular acceleration at disk A is αa= 9.375 rad/s^2 is acting in the anti-clockwise direction.

For Disk B,

T= 1/2Mar^2a

T= 1/2(3.5) (0.06)^2

=0.0063kg-m^2

We have,

Summation Mb= Summation(Mb) eff

Frb= Ibαb

(3N)(0.06m)= (0.0063kg-m^2) α

αb= 28.57 rad/s^2

B) ( ωa)o= 360rpm(2 pi/60)

= 1 pi rad/s

The disk will stop sliding where

ωara=ωbrb

(ωao-at)ra=αtr

(12pi-9.375t) (0.08)=28.57t(0.06)

(37.704-9.375t)(0.08)= 1.7142t

3.01632-0.75t= 1.7142t

t=1.22s

Now,

ωa=(ωa)o- t

12pi - 9.375(1.22)

37.704-11.4375

=26.267 rad/s

26.267× (60/2pi)

= 250.80

251rpm

The angular velocity at a, ω= 251rpm

Now,

ωb= αbt

=28.57(1.22)

=34.856rad/s

34.856rad/s × (60/2pi)

=332.807

= 333rpm

Therefore the angular velocity at b ω=333rpm

4 0
3 years ago
Read 2 more answers
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