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diamong [38]
3 years ago
10

If water is at 50* C and energy is removed from the water until it is at - 10* C

Physics
1 answer:
aliya0001 [1]3 years ago
3 0

Answer: B) Solid water to liquid water.

Explanation: This would be the melting of water or a transition of solid water to liquid water as energy is added.

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The superscripts in a chemical formula are the overall charge of an ion

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As an electromagnetic wave travels, what is the relationship between the
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Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c
Travka [436]

Answer:

22.1 V

Explanation:

We are given that

A_1=0.7 cm^2=0.7\times 10^{-4} m^2

A_2=2.5 cm^2=2.5\times 10^{-4} m^2

A_3=2.2 cm^2=2.2\times 10^{-4} m^2

A_4=3 cm^2=3\times 10^{-4} m^2

Using 1cm^2=10^{-4} m^2

We know that

R=\frac{\rho l}{A}

In series

R=R_1+R_2+R_3+R_4

R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}

R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}

Substitute the values

R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})

R=\rho l(2.62\times 10^4)

V=145 V

I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}

Voltage across the 2.5 square cm wire=IR=I\times \frac{\rho l}{A_2}

Voltage across the 2.5 square cm wire=\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V

Voltage across the 2.5 square cm wire=22.1 V

6 0
3 years ago
A parallel beam of light containing orange (610 nm) and violet (410 nm) wavelengths goes from fused quartz to water, striking th
Fudgin [204]

Answer:

0.98°

Explanation:

First we find the refracting angle of orange in water

Which is given as

စr= sin^-1. ( 1.456 sin60°/1.33)

= 71.2°

Then that of violet in water

စv=sin^-1. ( 1.468sin60°/1.342)

= 72.3°

So angle between boths colours is the difference

71.2- 72.3= 0.98°

6 0
4 years ago
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