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:
C) 2.44 × 106 N/C
Explanation:
The electric flux through a circular loop of wire is given by
where
E is the electric field
A is the cross-sectional area
is the angle between the direction of the electric field and the normal to A
The flux is maximum when , so we are in this situation and therefore
, so we can write
Here we have:
is the flux
d = 0.626 m is the diameter of the coil, so the radius is
r = 0.313 m
and so the area is
And so, we can find the magnitude of the electric field: