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LenaWriter [7]
3 years ago
12

A fuel cell that is becoming very popular for cars runs on:

Physics
1 answer:
jok3333 [9.3K]3 years ago
7 0
It runs on Hydrogen gas.

Actually, using hydrogen as a fuel is not new. We used to use it on air vehicle like air balloon. But back then, we still cannot figure out how to safely use this because Hydrogen exlodes rather easily
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A crate filled with delicious junk food rests on a horizontal floor. Only gravity and the support force of the floor act on it,
stealth61 [152]

The words "... as shown ..." tell us that there's a picture that goes along
with this question, and you decided not to share it.  That's sad and
disappointing, but I think the question can be answered without seeing
the picture.

The net force on the crate is zero.  Evidence for this is that fact that
the crate is just sitting there.  If the net force on an object is not zero,
then the object is accelerating ... it's either speeding up, slowing down,
or its the direction of its motion is changing.  If none of these things is
happening, then the net force on the object must be zero.

4 0
3 years ago
A stack of 55 business cards is 1.85 cm tall. Use this information to determine the thickness of one business
Sindrei [870]
To find the answer, take 55 and divide it by 1.85 to get the thickness of one card. In this case the answer would be 29.72973 cm. each.
8 0
3 years ago
Read 2 more answers
All the planets orbit the Sun
aivan3 [116]

Answer:

Yes, they do

Explanation:

7 0
3 years ago
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What is nuclear energy
irina1246 [14]
Nuclear energy <em>is t</em><span><em>he energy released during nuclear fission or fusion.

</em>
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6 0
3 years ago
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Determine the wavelength of incident electromagnetic radiation required to cause an electron transition from the n = 5 to the n
ollegr [7]
The correct answer is: wavelength = 4562 nm

Explanation:

Rydberg's formula is given as:
\frac{1}{\lambda} = R[ \frac{1}{n_1^2}  - \frac{1}{n_2^2} ] --- (1)

Where 
R = Rydberg's constant = 1.096 * 10^7 per meter
n_1 = 5
n_2 = 7

λ = Wavelength

Plug in the values in (1):

(1)=> \frac{1}{\lambda} = (1.096 * 10^7)[ \frac{1}{5^2} - \frac{1}{7^2} ]

\frac{1}{\lambda} = (1.096 * 10^7)[ 0.04 - 0.020 ] \\ \lambda =  \frac{1}{(1.096 * 10^7)[0.020 ]} \\ \lambda = 4562 nm
8 0
3 years ago
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