Answer:
Final velocity of electron, 
Explanation:
It is given that,
Electric field, E = 1.55 N/C
Initial velocity at point A, 
We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :
........(1)
a is the acceleration, 
We know that electric force, F = qE
Use above equation in equation (1) as:
v = 647302.09 m/s
or
So, the final velocity of the electron when it reaches point B is 
. Hence, this is the required solution.
Answer:
F = - 3.56*10⁵ N
Explanation:
To attempt this question, we use the formula for the relationship between momentum and the amount of movement.
I = F t = Δp
Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say
v = d / t
t = d / v
Given that
m = 26 g = 26 10⁻³ kg
d = 50 mm = 50 10⁻³ m
t = d/v
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
F t = m v - m v₀
This is so, because the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵
F = - 3.56*10⁵ N
The negative sign is as a result of the force exerted against the bullet
Ooh. You don't knock on your own hotel door- <span> a knock at the door is your key word.
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I'm pretty sure it would be A. As fluids are able to move and flow pretty easily.
This means acceleration a is constant.
Let
a) vo be the initial speed, at t=0
b) v be the final speed after time t
c) d distance travelled in time t
Then we have:
a) v=vo+a×t
b) v²=vo²+2×a×d (Galilei's equation)
c) d=vo×t+a×t²/2
d) average speed vm=(vo+v)/2
