Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
Answer:
0.1g to 0.0000001g hope it helps uu
Answer:
So airplane will be 1324.9453 m apart after 2.9 hour
Explanation:
So if we draw the vectors of a 2d graph we see that the difference in angles is = 83 - 44.3 = 
Distance traveled by first plane = 730×2.9 = 2117 m
And distance traveled by second plane = 590×2.9 = 1711 m
We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.
Using the law of cosine,
representing the distance between the planes, we see that:

d = 1324.9453 m
Answer:
e.26m/s
Explanation:
Vf=Vi+at (1)
Vf=9j+(2i-4j)t
X= X₀+at
now, in the i direction
15=O+2t or t=7.5 when x position is 15
Lets put that into the (1) equation, solve for Vf.
Vf=9j+(2i-4j)7.5
Vf= 15i - 21j
Speed=
Vf= 25.8 m/s