1 Point
1 answer:
You might be interested in
Answer:
4.0 m/s
Explanation:
In the first part of the run, the athlete runs a distance of
at a speed of
So, the time he/she takes is
In the second part of the run, the athlete covers an additional distance of
with a speed
So, the time taken in this second part is
So, the total distance covered is
d = 300 m + 300 m = 600 m
And the total time taken
t = 100 s + 50 s = 150 s
Therefore, the average speed for the entire trip is
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
#A
Mass=4.4kg
acceleration=-1.74m/s^2
Use newtons second law
#B
initial velocity=u
Final velocity=v=0
Acceleration=a=-1.74m/s^2
Time=t=1.27s