2. Do the results by using a larger sample demonstrate the advantage of having a larger sample in the scattering experiment? Exp
1 answer:
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Answer:
30.22 hours
Explanation:
Given data:
A= l² = (2 x )² = 4 x
m²
Length 'L' = 5m
current '' = 2 A
density of free electrons 'n'= 8.5 x /m³
Current Density 'J' = / A
J= 2/4 x
J= 5 x A/m²
We can determine the time required for an electron to travel the length of the wire by
T= L/ Vd
Where,
L is length and Vd is drift velocity.
Vd can be defined by J/ n|q|
where,
n is the charge-carrier number density
|q| is is the charge carried by each charge carrier
=>1.6 x C
T= L/ Vd
Therefore,
T= L . n|q| / J
T= (4 x 8.5 x x |1.6 x
|)/5 x
T= 108800 seconds =>1813.33 minutes
Converting minute into hours:
T= 30.22 hours
Thus, time that is required for an electron to travel the length of the wire is 30.22 hours
Answer:
Average speed of car in the first trip is 10 km/hr
Explanation:
It is given that first the car drives 6 hours to the east
Then travels 12 km to west in 3 hours
Average speed for the entire trip = 8 km/hr
Total time = 3+6 = 9 hour
So distance traveled in 9 hour = 9×8 = 72 km
As the car travel 12 km in west so distance traveled in east = 72-12 = 60 km
Time by which car traveled in east = 6 hour
So speed
So average speed of car in the first trip is 10 km/hr