Answer:
To maintain enough time to prevent a collision, a system operating in air traffic where aircraft speed does not
fall below 100 km/h (most medium-sized UAVs and GA aircraft) will need to be able to detect obstacles which
subtend an arc-width of as small as 0.125 mra
It would have to be c because it is a chemical change. your welcome!!!!!
Answer:
If all these three charges are positive with a magnitude of
each, the electric potential at the midpoint of segment
would be approximately
.
Explanation:
Convert the unit of the length of each side of this triangle to meters:
.
Distance between the midpoint of
and each of the three charges:
Let
denote Coulomb's constant (
.)
Electric potential due to the charge at
:
.
Electric potential due to the charge at
:
.
Electric potential due to the charge at
:
.
While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.
Hence, the electric field at the midpoint of
due to all these three charges would be:
.
A technique in which the muscles are stretched by an outside force is called Passive Stretching
The car is accelerating at 3 m/s² in the positive direction (to the right). By Newton's second law, the net force on the car in this direction is
∑ F = F[a] - F[f] - F[air] = ma
3100 N - 200 N - F[air] = (650 kg) (3 m/s²)
Solve for F[air] :
F[air] = 3100 N - 200 N - (650 kg) (3 m/s²)
F[air] = 3100 N - 200 N - 1950 N
F[air] = 950 N