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BabaBlast [244]
3 years ago
8

During the rainy season, we can observe lighting in the sky. Due to lighting, the atmospheric

Physics
1 answer:
alina1380 [7]3 years ago
6 0

Answer:

A chemical change has taken place to form nitric oxide.

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A sophomore with nothing better to do adds heat to a mass 0.300 kg of ice at 0.0 âc until it is all melted.
kramer
Melting ice?  Man, that's cold
8 0
3 years ago
"In a pure substance all the particles must be identical; therefore pure substances are composed only of elements." Do you agree
kenny6666 [7]

I do not agree with the statement.
The "substance" can be a compound.  It's "pure"
as long as there's nothing else in it but its name.

'Pure' water is 100% H₂O with nothing else in it.
'Pure' table salt is 100% NaCl with nothing else in it.
'Pure' carbon dioxide is 100% CO₂ with nothing else in it.

These example substances are all compounds, not elements.
 
6 0
3 years ago
Enter an expression in the box to Write the equation of the line perpendicular to y=-3x-1 that passes through the point (3,4).
Dafna11 [192]

Answer: -3x+13

Explanation:

4 0
3 years ago
In the diagram, the trough of the wave is shown by:
Kisachek [45]

Answer:

the correct representation of the trough is b

3 0
3 years ago
An isolated conducting sphere has a 10 cm radius. One wire carries a current of 1.000 002 0 A into it. Another wire carries a cu
romanna [79]

Answer:

t = 5.56 ms

Explanation:

Given:-

- The current carried in, Iin = 1.000002 C

- The current carried out, Iout = 1.00000 C

- The radius of sphere, r = 10 cm

Find:-

How long would it take for the sphere to increase in potential by 1000 V?

Solution:-

- The net charge held by the isolated conducting sphere after (t) seconds would be:

                                   qnet = (Iin - Iout)*t

                                   qnet = t*(1.000002 - 1.00000) = 0.000002*t

- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:

                                   V = k*qnet / r

Where,                        k = 8.99*10^9   ..... Coulomb's constant

                                   qnet = V*r / k

                                   t = 1000*0.1 / (8.99*10^9 * 0.000002)

                                   t = 5.56 ms

                                   

7 0
3 years ago
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