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Ostrovityanka [42]

2 years ago

10

The refractive index of water is greater than that of air but is less than that of glass or a diamond. Under which condition can

there be total internal reflection in water?

1 answer:

Xelga [282]2 years ago

5 0

For total internal reflection to happen, light must travel from a medium object of higher density to a medium of lower density. To make sure that total internal reflection will happen in water, then the light must originate from glass or diamond before entering water.

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Jesse wants to know how well a particular brand of car wax protects his car from dirt. What is the independent variable ?

miv72 [106K]

The brand of car wax

4 0

2 years ago

Unit 2 Lesson 9: Temperature & Heat Unit Test does anyone have the answers

IRISSAK [1]

Explanation:

where is the question

I did not understood this question

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2 years ago

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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

2 years ago

What is the net displacement of the particle between 0 seconds and 80 seconds?

Illusion [34]

After a thorough research, there exists the same question that has choices and the link of the graph (http://i37.servimg.com/u/f37/16/73/53/52/graph410.png)

<span>Choices:
A. 160 meters
B. 80 meters
C. 40 meters
D. 20 meters
E. 0 meters
</span>
The correct answer is letter E. 0 meters.

7 0

2 years ago

The electric field in a particular thundercloud is 3.8 x 105 N/C. What is the acceleration (in m/s2) of an electron in this fiel

Svetradugi [14.3K]

Answer:

Answer:

6.68 x 10^16 m/s^2

Explanation:

Electric field, E = 3.8 x 10^5 N/C

charge of electron, q = 1.6 x 10^-19 C

mass of electron, m = 9.1 x 10^-31 kg

Let a be the acceleration of the electron.

The force due to electric field on electron is

F = q E

where q be the charge of electron and E be the electric field

F = 1.6 x 10^-19 x 3.8 x 10^5

F = 6.08 x 10^-14 N

According to Newton's second law

Force = mass x acceleration

6.08 x 10^-14 = 9.1 x 10^-31 x a

a = 6.68 x 10^16 m/s^2

Explanation:

8 0

2 years ago