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Nataliya [291]
2 years ago
7

Convert.46 to a percentage

Engineering
1 answer:
natima [27]2 years ago
6 0

Answer:

46%

Explanation:

Move the decimal two times and you should have a precentage.

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The particle travels along the path defined by the parabola y=0.5x2, where x and y are in ft. If the component of velocity along
JulsSmile [24]

Answer:

D=41.48 ft

a=54.43\ ft/s^2

Explanation:

Given that

y=0.5 x²                      

Vx= 2 t

We know that

V_x=\dfrac{dx}{dt}

At t= 0 ,x=0  

x=\int V_x.dt

At t= 3 s

x=\int_{0}^{3} 2t.dt

x=[t^2\left\right ]_0^3

x= 9 ft

When x= 9 ft then

y= 0.5 x 9²  ft

y= 40.5 ft

So distance from origin is

x= 9 ft ,y= 40.5 ft

D=\sqrt{9^2+40.5^2} \ ft

D=41.48 ft

a_x=\dfrac{dV_x}{dt}

Vx= 2 t

a_x= 2\ ft/s^2

At t= 3 s , x= 9 ft

y=0.5 x²    

a_y=\dfrac{d^2y}{dt^2}

y=0.5 x²    

\dfrac{dy}{dt}=x\dfrac{dx}{dt}

\dfrac{d^2y}{dt^2}=\left(\dfrac{dx}{dt}\right)^2+x\dfrac{d^2x}{dt^2}

Given that

\dfrac{dx}{dt}=2t

\dfrac{dx}{dt}=2\times 3

\dfrac{dx}{dt}=6\ ft/s

a_y=\dfrac{d^2y}{dt^2}=6^2+9\times 2\ ft/s^2

a_y=54\ ft/s^2

a=\sqrt{a_x^2+a_y^2}\ ft/s^2

a=\sqrt{2^2+54^2}\ ft/s^2

a=54.43\ ft/s^2

7 0
3 years ago
The minimum requirements for engineering documents are enumerated in
vladimir1956 [14]

Answer:

The answer will be Rule 61G15-23 F.A.C, relating to Seals.

Explanation:

According to the description given by: Florida administrative code&Florida administrative register the Minimum requirements for engineering documents are in the section 'Final 61G15-23'  from 11/3/2015. This document provides specifications of materials required for the safe operation of the system that is the result of engineering calculations, knowledge and experience.

8 0
3 years ago
Which of the following devices would you expect to consume the most energy for each hour that it operates?
Gennadij [26K]
An electric space heater
4 0
2 years ago
A certain part of the cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a di
Bezzdna [24]

Answer :

<h3>Flow rate in pipe B is = 0.3094 \frac{m^{3} }{s}</h3>

Explanation:

Given :

Length of pipe A L_{A}  = 1500 m

Length of pipe B L_{B} = 2500 m

Flow rate through pipe A Q_{A}  = 0.4 \frac{m^{3} }{s}

Diameter of pipe D = 30 \times 10^{-2} m

Velocity from pipe A,

  V _{A} = \frac{Q_{A} }{A}

  V _{A} = \frac{0.4 \times 4 }{\pi ( 30 \times 10^{-2} )^{2}  }

  V_{A}  = 5.66 \frac{m}{s}

Here, head loss is same because height is same.

    h_{a} = h_{b}

L_{A} V_{A} ^{2} = L_{B}  V_{B} ^{2}

V_{B} = \sqrt{\frac{1500}{2500}}    (5.66)

V_{B} = 4.38 \frac{m}{s}

Now rate of flow from pipe B is,

Q_{B}  = V_{B} A

Q_{B}  = \frac{\pi }{4}  (0.3)^{2} \times 4.38

Q_{B} = 0.3094 \frac{m^{3} }{s}

4 0
3 years ago
One kilogram of "as received" yard trimmings is made up of approximately 620 g moisture, 330 g of decomposable organic matter (r
Evgesh-ka [11]

Answer:

Explanation:

(a) Given that 620g moisture and 330g decomposable organic matter in yard trimming is represented by C₁₂.₇₆H₂₁.₂₈O₉.₂₆N₀.₅₄

Given the atomic mass of Carbon C = 12, Hydrogen H = 1, Oxygen O = 16 and Nitrogen N = 14

1 mole of trimming = 12*12.76 + 1*21.28 + 16*9.26 + 14*0.54

=  153.12 + 21.28 + 148.16 + 7.56

= 330.12 g/mol

which means 1 kg of as received trimming has 330 g of decomposable that produce 1 mole of decomposable

The moles of methane produced will be given as

m = (4a + b -2c - 3d)/8

= (4*12.76 + 21.28 - 2*9.26 - 3*0.54)/8

= (51.04 + 21.28 - 18.52 - 1.62)/8

= 52.18/8

= 6.5225

(b) Volume of methane V is given as

V = (0.0224 m³ CH₄mol/CH₄) × (6.5225 mol CH₄/ kg)

= 0.1461 m³ CH₄/kg lawn trimmings

(c) Energy will be given as

CH₄Energy = 6.5225 mol of CH₄/kg × 890 kJ/mol

= 5805.025

≈ 5805 kJ/kg

5 0
2 years ago
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