A client has developed dystrophic calcification as a result of macroscopic deposition of calcium salts. The tissue that would be
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Recall that to compute for the emf of a circuit given current and inductance, we must recall that
where I is the current (A), M is the mutual inductance (h), and t is the time (ms). Since the current must not exceed 80.0 V, we have
From this, we see that it must take at least 0.35 ms so it doesn't exceed 80 V.
Answer: 0.35 ms
where I is the current (A), M is the mutual inductance (h), and t is the time (ms). Since the current must not exceed 80.0 V, we have
From this, we see that it must take at least 0.35 ms so it doesn't exceed 80 V.
Answer: 0.35 ms
Answer:
Diagrams in pictures
Explanation:
Using energy I can get
m g h = 1/2 m v^2
So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.
(H= 1,3m sin25)
V=2,32m/s
V= a t
And
X= v t +1/2 a t^2
Knowing v=2,32 m/s and x= 1,3 m
I can get
a= 6,21m/s2
F= m a
I can get the force of the box when it collides with the spring
F= 12, 425 N
The force the spring makes on the box then is
F = -12,425N = -k d
Then the spring's constant is k= 51,75N/m
To make the two diagrams I need the functions of time when the box slows down
I use the same two equations
V= a t
And
X= v t + 1/2 a t^2
Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.
I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).
Then,
V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec
X(t)= 2,32 m/s t + 1/2 33,67 m/s2 t^2.
for time between 0 and 0,689 sec
Diagrams and equations are in the pictures
