I’m it’s about 3.4878 energy in joules
A ) v = v o + a t ( the acceleration will be negative )
9.50 = 16.0 + a * 1.2
a * 1.2 = -16.0 + 9.50
a * 1.2 = - 6.5
a = - 6.5 : 1.2
a = - 5.4167 m/s²
F = m * a = 950 kg * 5.4167 m/s²
F = 5,145.8 N ( the average force exerted on a car during braking )
b ) d = v o - a t² / 2
d = 16.0 * 1.2 - ( 5.4167 * 1.2² / 2 ) =
= 19.20 - 3.90 = 15.30 m
If Earth's axis was "straight up and down" instead of tilted, then ...
<span>-- There would be no seasons.
-- The climate at any one place would be the same all year around.
-- The days would be the same length, everywhere,
and all year around.
-- So would the nights.
-- The sun would be up a little more than 12 hours every day.
It would be down a little less than 12 hours every day.
-- At the middle of the day, the sun would be at the same height
in the sky all year around, not higher in some months and lower
in others.
-- The equator would be the only place on Earth where the sun
could ever be directly over your head.
-- If you were at the north pole or the south pole, the sun would be
down on the horizon, and it would just go around and around you
every day. It would never rise or set, and it would never get any
higher or lower.
</span>
Answer:
E=12.2V/m
Explanation:
To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.
The equation is given by,
![V=\frac{I}{nAq}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BI%7D%7BnAq%7D)
Where,
V= Drift Velocity
I= Flow of current
n= number of electrons
q = charge of electron
A = cross-section area.
For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is
![\frac{I}{q} = 1.2*10^{18}](https://tex.z-dn.net/?f=%5Cfrac%7BI%7D%7Bq%7D%20%3D%201.2%2A10%5E%7B18%7D)
![A= 1.3*10^{-8}m^2](https://tex.z-dn.net/?f=A%3D%201.3%2A10%5E%7B-8%7Dm%5E2)
![n=6.3*10^{28} e/m^3](https://tex.z-dn.net/?f=n%3D6.3%2A10%5E%7B28%7D%20e%2Fm%5E3)
Mobility
We can find the drift velocity replacing,
![V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1.2%2A10%5E%7B18%7D%7D%7B%281.3%2A10%5E%7B-8%7D%29%286.3%2A10%5E%7B28%7D%29%7D)
![V= 1.465*10^-3m/s](https://tex.z-dn.net/?f=V%3D%201.465%2A10%5E-3m%2Fs)
The electric field is given by,
![E= \frac{V}{\omicron{O}}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7BV%7D%7B%5Comicron%7BO%7D%7D)
![E=\frac{1.465*10^-3}{1.2*10^{-4}}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1.465%2A10%5E-3%7D%7B1.2%2A10%5E%7B-4%7D%7D)
![E=12.2V/m](https://tex.z-dn.net/?f=E%3D12.2V%2Fm)