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worty [1.4K]
2 years ago
13

A person uses a match to light charcoal in a grill. Which statement describes

Physics
1 answer:
dolphi86 [110]2 years ago
6 0

Answer:

It is flammable.

Explanation:

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In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point
jarptica [38.1K]
The difference in electric potential energy between the two points is
\Delta U = q \Delta V
where q is the magnitude of the charge and \Delta V is the electric potential difference.

But for energy conservation, the difference in electric potential energy \Delta U between the two points is equal to the work done to move the charge between A and B:
W=\Delta U
so we have
W=q \Delta V

and by substituting the numbers of the problem, we find the value of \Delta V:
\Delta V =  \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V
3 0
2 years ago
A car of 1400 kg is subject to multiple forces which produce an acceleration of 3.5 m/s2 directed north. Find the net force.​
Greeley [361]

Answer:

will

Explanation:

3 0
2 years ago
What are valence electrons?
SashulF [63]

Answer:

The correct answer is D. Electrons in an atom that can bond with other atoms.

Explanation:

For those of you that need it still

8 0
2 years ago
A Michelson interferometer uses red light with a wavelength of 656.45 nm from a hydrogen discharge lamp. Part A How many bright-
kati45 [8]

Answer:

51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7cm

Explanation:

The number of maxima appearing when the mirror M moves through distance \Delta L is given as follows,

\Delta m = \frac{\Delta L}{\frac{\lambda}{2}}

Here,

\Delta L= is the distance moved by the mirror M

\lambda is the wavelenght of the light used.

\Delta L= 0.017m

\lambda = 656.45*10^{-9}m

\Delta m = \frac{0.017}{\frac{656.45*10^{-9}}{2}}

\Delta m = 51793.72

Therefore, 51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7

3 0
2 years ago
Hydrogen line spectrum lies entirely within visible range
mart [117]

No, that's silly.

You've got your Pfund series where electrons fall down to the 5th level,
your Brackett series where they fall to the 4th level, and your Paschen
series where they fall to the 3rd level.  All of those transitions ploop out
photons at Infrared wavelengths.

THEN next you get your Balmer series, where the electrons fall in
to the 2nd level.  Most of those are at visible wavelengths, but even
a few of the Balmer transitions are in the Ultraviolet.

And then there's the Lyman series, where electrons fall all the way
down to the #1 level.  Those are ALL in the ultraviolet. 
6 0
3 years ago
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