A 17,000-kg airplane lands with a speed of 82 m/s on a stationary aircraft carrier deck that is 115 m long. find the work done b
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The change in potential energy of the proton is 5.6 x Joule
A uniform electric field is where the electric field strength is the same at all points in the field. In the uniform field, the force experienced by a charge is the same no matter where it is placed in the field.
Given that a proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.
- The distance d = 10 cm = 0.1 m
- Electric field E = 3.5 KN/C
- Proton charge q = 1.6 x
C
The Work done = Fd
but F = Eq
Recall that Electric field E = F/q = V/d
Where V = potential difference.
Let us first calculate the V
E = V/d
V = Ed
Substitute all the parameters into the formula above
V = 3.5 × 10³ × 0.1
V = 350 v
from F/q = V/d
make F the subject of formula and substitute it in work formula
F = Vq/d
W.D = Vq/d x d
W.D = Vq
Substitute all the parameters into the formula above
W.D = 350 x 1.6 x
W.D = 5.6 x J
Work done = Energy = Potential Energy
Therefore, the change in potential energy of the proton is 5.6 x <em> Joule</em>
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Answer:
a)
b)
Explanation:
Given data:
Electric field = 1.47 N/C
velocity of electron is
distance of point b from point A is 0.55 m
we know that acceleration of particle is given as
a) for electron
from equation of motion we have
b) for proton
from equation of motion we have