Acceleration= change in velocity/time; 5/2.8 , so a=1.785714286
The work done on the sail is 600 J
Explanation:
The work done to lift the sail is equal to the gain in gravitational potential energy of the sail, therefore is:
where
m is the mass of the sail
g is the acceleration of gravity
(mg) is the weight of the sail
is the change in height of the sail
In this problem we have
mg = 150 N (weight)
Substituting, we find the work done:
Learn more about work:
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Answer:
1)
a) f = 1m × 2 × (5A / √2) × (5A / √2) / 0.003m = 0.00166... (66 is repeating)
b) The currents on two wires on a AC chord are always moving in opposite direction and so they are always replusing.
c) There needs to be a sheath to dampen the replusing, fluctuating force of the wires.
2)
a) v = √( ( (-2)(-1.6 × 10^(-16))(3000V) ) / (2.84 × 10^(-20)kg) ) = 5.81227 × 10^3
b) Any ion transversing a chamber having a magnetic field will deflect.
c) The direction of the electric field is vertical because it's perpendicular to the plates. The electric field magnitude is independent from the magnitude of the magnetic field and charge. So it's not possible to find the magnitude of the electric field, without knowing the voltage on the plates, the distance between the plates, and the dielectric constant.
d) Assuming the mangetic field remained, the path of the negative ions will be deflected vertically given that the magnetic field is horizontally perpendicular to the negative charged ions movement.
Sorry it took so long :) If anything is incorrect please let me know.
