Explanation:
Answer:
Magnetic field, B = 0.016 Tesla
Explanation:
It is given that,
Velocity of electron, v=9.6\times 10^5\ m/sv=9.6×10
5
m/s
Magnetic force, F=2.6\times 10^{-15}\ NF=2.6×10
−15
N
Charge, q=1.6\times 10^{-19}\ Nq=1.6×10
−19
N
The magnetic force is given by :
F=qvBF=qvB
B=\dfrac{F}{qv}B=
qv
F
B=\dfrac{2.6\times 10^{-15}}{1.6\times 10^{-19}\times 9.6\times 10^5}B=
1.6×10
−19
×9.6×10
5
2.6×10
−15
B=0.016\ TB=0.016 T
So, the magnitude of magnetic field is 0.016 Tesla. Hence, this is the required solution.
Answer:
what are the options for me
Change in speed = (speed at the end) minus (speed at the beginning)
= (0 - 20 m/s) = -20 m/s
Acceleration = (change in speed) / (time for the change)
= (-20 m/s) / (3.5 sec) = -5.71 m/s²
Notice that this story probably didn't happen on Earth, because
the acceleration due to gravity on Earth is 9.8 m/s².
More convenient
More commonly used
"Lightyear" is a <em>distance</em> ... specifically, the distance that light travels
through vacuum in one year of 365 days.
It's equivalent to 9.454 x 10^12 kilometers, or 5.875 x 10^12 miles .
