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attashe74 [19]
4 years ago
8

Energy captured during the ""photo"" part of photosynthesis is stored in _____ during the ""synthesis"" part of the process.

Physics
1 answer:
andre [41]4 years ago
5 0

Energy captured during the ""photo"" part of photosynthesis is stored in  <u>covalent bond</u> during the ""synthesis"" part of the process.

<u>Explanation:</u>

When carbon dioxide, water and sunlight are combindly processed by Plants, algae and a set of bacteria called cyanobacteria to become photoautotrophs, then the process goes is named as Photosynthesis. It generates oxygen, Glyceraldehyde-3-phosphate (G3P), common high-energy carbohydrate molecules which result into glucose, sucrose or other sugar molecules which comprises covalent energy-saving bonds.

Thus the species breakdown these molecules to exhibit energy for cellular functioning. In light-dependent processes, chlorophyll absorbs the radiation from the sunlight and converts it into chemical energy in the form of electron carrier derivatives such as ATP and NADPH. Carbohydrate molecules are constructed from carbon dioxide in light-independent processes i.e in the Calvin cycle, using the chemical energy obtained throughout the light-dependent processes.

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C - I’m pretty sure!
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3 years ago
Assume you need to design a hydronic system that can deliver 80,000 Btu/hr. What flow rate of water is required if the temperatu
PolarNik [594]

Answer:

At 10°F change in temperature

Mass flowrate = 1.01 kg/s = 2.227 lbm/s

Volumetric flowrate = 1010 m³/s = 35667.8 ft³/s

At 20°F change in temperature

Mass flowrate = 0.505 kg/s = 1.113 lbm/s

Volumetric flowrate = 505 m³/s = 17833.9 ft³/s

Explanation:

80000 btu/hr = 23445.7 W

P = ṁc(ΔT)

ṁ = MASS flowrate

c = specific heat capacity of water = 4182 J/kg.K,

ΔT = change in temperature = 10°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 10°F = 10×10/18 = 5.556°C = 5.556K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 5.556)

ṁ = 23445.7/(4182 × 5.556)

ṁ = 1.01 kg/s = 2.227 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 1.01 = 1010 m³/s = 35667.8 ft³/s

For a change of 20°F,

ΔT = change in temperature = 20°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 20°F = 20×10/18 = 11.1111°C = 11.111K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 11.111)

ṁ = 23445.7/(4182 × 11.111)

ṁ = 0.505 kg/s = 1.113 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 0.505 = 505 m³/s = 17833.9 ft³/s

Hope this Helps!!!

4 0
3 years ago
Which stars has the coolest surface temperature?
Delicious77 [7]

Answer:

Red Dwarf

Explanation:

  • Red Dwarf have low temperature having range from 2500 K to 4000 K.
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  • Proxima Centauri is a red Dwarf.
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  • They have small mass and have high density.
5 0
3 years ago
In a physics experiment, a ball is released from rest, and it falls toward the ground. The timer was not paying attention but es
tigry1 [53]

Answer:

(A) –14m/s

(B) –42.0m

Explanation:

The complete solution can be found in the attachment below.

This involves the knowledge of motion under the action of gravity.

Check below for the full solution to the problem.

4 0
4 years ago
If a rock climber accidentally drops a 77.5-g piton from a height of 215 meters, what would its speed be just before striking th
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Let us list out what we know from the question.

Initial Velocity V_{i}  = 0 since the piton is 'dropped'.

Vertical Displacement of the piton D = 215 m

Acceleration due to gravity a = 9.8 m/s^{2}

Final Velocity V_{f} = ?

Using the equation, V^{2} _{f} = V^{2} _{i}  + 2aD and plugging in the known values, we get

V^{2} _{f}  = 0^{2}  + 2(9.8)(215)

Simplifying by taking square-root on both sides gives us V_{f} =  64.915 m/s

Thus, the speed of the piton just before striking the ground is 65 m/s.

4 0
3 years ago
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