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Montano1993 [528]
3 years ago
11

3. To most people sugar is a white substance. Is this the sugar that plants use during photosynthesis?

Physics
2 answers:
ki77a [65]3 years ago
7 0

Answer:

I don't believe so...

Also... I have a question up for world history, if any of you mind checking it out, i rlly need some help with it. thanks so much and good luck on your exams

Explanation:

drek231 [11]3 years ago
4 0

Answer:

No

Explanation:

,I personally don't think so

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A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com
bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature =20^{\circ}C

Final Temperature =80^{\circ}C

mass flow rate of cold fluid \dot{m_c}=1.2 kg/s

Initial Geothermal water temperature T_h_i=160^{\circ}C

Let final Temperature be T

mass flow rate of geothermal water \dot{m_h}=2 kg/s

diameter of inner wall d_i=1.5 cm

U_{overall}=640 W/m^2K

specific heat of water c=4.18 kJ/kg-K

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)

2\times (160-T)=1.2\times (80-20)

160-T=36

T=124^{\circ}C

As heat exchanger is counter flow therefore

\Delta T_1=160-80=80^{\circ}C

\Delta T_2=124-20=104^{\circ}C

LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}

LMTD=\frac{80-104}{\ln \frac{80}{104}}

LMTD=91.49^{\circ}C

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

\dot{m_c}c(80-20)=U\cdot A\cdot (LMTD)

A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2

A=\pi DL=5.144

L=\frac{5.144}{\pi \times 0.015}

L=109.16 m

6 0
3 years ago
The nuclei of large atoms, such as uranium, with 9292 protons, can be modeled as spherically symmetric spheres of charge. The ra
Scrat [10]

Answer:

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

Explanation:

Given

Number of protons = 92

Radius of nucleus r_n = 7.4 * 10^{-15} m

Distance of the electrons r_1 = 1.0 * 10^ {-10} m

Part 1

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21}  N/C

Part 2

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13}  N/C

Part 3

The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other

hence, the solution is

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

7 0
3 years ago
Vertical Motion
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3 0
3 years ago
An astronaut is taking a space walk near the shuttle when her safety tether breaks. what should the astronaut do to get back to
sergeinik [125]
The best way in handling in this situation is that in order for the astronaut to be able to get back to the shuttle is that he or she should take an object from his or her tool belt and to be thrown out away from the shuttle. This will allow her to weight lightly and safely return to the shuttle and would be easier for his or her to do so.
3 0
3 years ago
The efficiency of conversion of glucose to ATP during glycolysis is approximately:
liraira [26]

Answer: approximately 50%

Explanation:

4 0
2 years ago
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