His velocity is 3 m/s in the direction in which he is running in. which.
Answer:
86.51° North of West or 273.49°
Explanation:
Let V' = velocity of boat relative to the earth, v = velocity of boat relative to water and V = velocity of water.
Now, by vector addition V' = V + v'.
Since v' = 6.10 m/s in the north direction, v' = (6.10 m/s)j and V = 100 m/s in the east direction, V = (100 m/s)i. So that
V' = V + v'
V' = (100 m/s)i + (6.10 m/s)j
So, we find the direction,Ф the boat must steer to from the components of V'.
So tanФ = 6.10 m/s ÷ 100 m/s
tanФ = 0.061
Ф = tan⁻¹(0.061) = 3.49°
So, the angle from the north is thus 90° - 3.49° = 86.51° North of West or 270° + 3.49° = 273.49°
I think the corect answer would be C. When a police officer receives information that you are speeding in your vehicle, she is using the frequency of the wave to measure the Doppler Effect. As the speed of a vehicle increases, the greater the change of the frequency of the waves would be transmitted to the radar guns which is being used by the police officer. Doppler effect is an effect that is observed in sound and light waves as these waves move away or to the direction of the observer. This is being used in many applications like in astronomy, weather balloons, the radar guns,and for underwater researches.
The question doesn't give us enough information to answer.
The answer depends on the mass of the object, how long the force
acts on the object, the OTHER forces on the object, and whether the
object is free to move.
-- If you increase the force with which you push on a brick wall,
the amount of work done remains unchanged, namely Zero.
-- If you push on a pingpong ball with a force of 1 ounce for 1 second,
the ball accelerates substantially, it moves a substantial distance, and
so the work done is substantial.
-- But if you push on a battleship, even with a much bigger force ...
let's say 1 pound ... and keep pushing for a month ... the ship accelerates
microscopically, moves a microscopic distance, and the work done by
your force is microscopic.