A particle is released as part of an experiment. Its speed t seconds after release is given by v (t )equalsnegative 0.4 t square
d plus 2 t, where v (t )is in meters per second. a) How far does the particle travel during the first 2 sec? b) How far does it travel during the second 2 sec?
1 answer:
Answer:
a) 2.933 m
b) 4.534 m
Explanation:
We're given the equation
v(t) = -0.4t² + 2t
If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.
See attachment for the calculations
The conclusion of the attachment will be
7.467 - 2.933 and that is 4.534 m
Thus, The distance it travels in the second 2 sec is 4.534 m
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Answer:
595391.482946 m/s
Explanation:
E = Energy = 1.85 keV
I = Current = 5.15 mA
e = Charge of electron =
t = Time taken = 1 second
m = Mass of proton =
Velocity of proton is given by
The speed of the proton is 595391.482946 m/s
Current is given by
Number of protons is
The number of protons is