ΔVl = L di/dt
i = i₀e -t/T
di/dt = i₀ × (-1/T) e -t/T
ΔVl = L× (-I/T i₀e -t/T
ΔVl = -L/T i₀e -t/T
b. 15mm, i₀ = 36mA, T = 1.1m
t= Os
ΔVl = 0,491V
C. t = 1ms
ΔVl = 0.198V
t = 2ms
ΔVl = 0.08V
E. t = ms
ΔVl = 0.032V
Answer:
lunar node in other words is either of the two orbital nodes of the Moon, that is, the two points at which the orbit of the Moon intersects the ecliptic. The ascending node is where the Moon moves into the northern ecliptic hemisphere, while the descending node is where the Moon enters the southern ecliptic hemisphere
Answer:
≈ 6.68 m/s
Explanation:
A suitable formula is ...
vf^2 -vi^2 = 2ad
where vi and vf are the initial and final velocities, a is the acceleration, and d is the distance covered.
We note that if the initial launch direction is upward, the velocity of the ball when it comes back to its initial position is the same speed, but in the downward direction. Hence the problem is no different than if the ball were initially launched downward.
Then ...
vf = √(2ad +vi^2) = √(2·9.8 m/s^2·1.0 m+(5 m/s)^2) = √44.6 m/s
vf ≈ 6.68 m/s
The ball hits the ground with a speed of about 6.68 meters per second.
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We assume the launch direction is either up or down.
Your answer would be that Hubble categorized galaxies by their shape.
Theater=sin^-1(30/80)
theater=22
F=400gsin22=150N
F=ma
150=400a
a=0.375ms^-2
v^2-u^2=2as
v^2-0=2x0.375x80
v=7.75ms^-1
