All categories

2 years ago

Do you think that biofuel is the only solution to the problem of reducing our

Physics

2 answers:

Bingel [31]2 years ago

6 0

Answer:

Explanation:

Absolutely not. We have three possibilities other than biofuel. There is solar fuel directly from the sun through the use of solar panels. There is also fission which involves "extracting" energy from splitting an atom. Fusion is not yet accomplished, but easily the most promising way of replacing fossil fuels. I'm not very familiar with fusion, but it can be grossly simplified as the opposite of fission. The reason why fusion would be effective is because it has the most potential for a net energy gain.

I am Lyosha [343]2 years ago

4 0

Answer:

f biofuel production and use reduces our consumption of imported fossil fuels, we may become less vulnerable to the adverse impacts of supply

Explanation:

You might be interested in

Look at a photograph of a fault. notice how the right side appears lower than the left side. this happens when pieces of crust a

dybincka [34]

I believe it’s divergent boundary but I might be wrong

5 0

3 years ago

Read 2 more answers

What type of rock can melt to form magma

navik [9.2K]

Igneous rock your welcome

3 0

3 years ago

a cruise ship travels directly toward the dock with a velocity of 15 m/s relative to the water. A passenger walks 3 m/s in the s

solong [7]

Answer:

Explanation:

i believe a if not c

4 0

2 years ago

A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p

alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

$F=qvB$

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

$qvB = \frac{mv^2}{r}$

which can be rewritten as

$v=\frac{qB}{mr}$

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

$\frac{2\pi r}{T}=\frac{qB}{mr}$

So, we get:

$T=\frac{2\pi m r^2}{qB}$

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) $a. f/10$

The frequency of revolution of a particle in uniform circular motion is

$f=\frac{1}{T}$

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

$f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}$