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yuradex [85]
3 years ago
15

The question “How do plants convert sunlight to energy?” best represents which of the following?

Engineering
1 answer:
OverLord2011 [107]3 years ago
5 0

Answer:

D

Explanation:

I would say this awnser because its the only one that makes sence to me

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A pumping test was made in pervious gravels and sands extending to a depth of 50 ft. ,where a bed of clay was encountered. The n
Vikki [24]

Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.

Explanation:

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3 years ago
what are some questions about simple machines that don't make me sound like i didn't pay attention in the lesson?
Margaret [11]

Answer:

what are simple machines?

Explanation:

it is 2020 let's be honest all

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3 years ago
The following program includes fictional sets of the top 10 male and female baby names for the current year. Write a program tha
otez555 [7]

Answer:

Define Variables and Use List methods to do the following

Explanation:

#<em>Conjoins two lists together</em>

all_names = male_names.union(female_names)

#<em>Finds the names that appear in both lists, just returns those</em>

neutral_names = male_names.intersection(female_names)

#<em>Returns names that are NOT in both lists</em>

specific_names = male_names.symmetric_difference(female_names)

3 0
3 years ago
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The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centr
Sedaia [141]

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

(\sigma_ x)  = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}

Since d = 2r

Then:

(\sigma_ x)  = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}

(\sigma_ x)  = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}

(\sigma_ x)  = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}

(\sigma_ x)  = 11917000.38

(\sigma_ x)  = 11.917 \times 10^6 \ Pa

(\sigma_ x)  = 11.917 \ MPa

\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\  \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

\tau _{xy} =0

3 0
3 years ago
A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m^3, undergoes a constant-pressure expansion at
NNADVOKAT [17]

Answer:

Q = E + W = 0.25 + 0.4 = 0.29KJ

W = 0.4KJ

b) 2000KJ

Explanation:

The concept applied here is the first law of thermodynamics ,i.e the equation of the first law.

mathematically from first law,

dE = dQ - dW

where E is internal energy

Where energy may be transfereed as eitherf work (W) or heat (Q)

Work on the other hand can be at constant volume and pressure.

at constant pressure, W = Integral (pdV), with final volume(Vf) as the upper limit and initial volume(Vi) as the lower limit

Work = p(Vf -Vi)

attempting the first question, Vi = 0.1metre cube, Vf = 0.12metre cube, p = 2bar, p(atm) = 1bar, E = U = 0.25KJ

from W = p(Vf -Vi) = 2 x 100000 ( 0.12 - 0.1) = 400N/m = 0.4KJ

To get the heat transfer, from dE = dQ - dW

Q = E + W = 0.25 + 0.4 = 0.29KJ

b) for the second question ; from Pdv = p(Vf -Vi), but the pressure here is in atmosphere

W = 100000 ( 0.12 - 0.1) = 2000KJ

6 0
4 years ago
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