Answer:
If the canoe heads upstream the speed is zero. And directly across the river is 8.48 [km/h] towards southeast
Explanation:
When the canoe moves upstream, it is moving in the opposite direction of the normal river current. Since the velocities are vector (magnitude and direction) we can sum each vector:
Vr = velocity of the river = 6[km/h}
Vc = velocity of the canoe = -6 [km/h]
We take the direction of the river as positive, therefore other velocity in the opposite direction will be negative.
Vt = Vr + Vc = 6 - 6 = 0 [km/h]
For the second question, we need to make a sketch of the canoe and we are watching this movement at a high elevation. So let's say that the canoe is located in point 0 where it is located one of the river's borders.
So we are having one movement to the right (x-direction). And the movement of the river to the south ( - y-direction).
Since the velocities are vector we can sum each vector, so using the Pythagoras theorem we have:
Answer:
car B will be 30 Km ahead of car A.
Explanation:
We'll begin by calculating the distance travelled by each car. This is illustrated below:
For car A:
Speed = 40 km/h
Time = 3 hours
Distance =?
Speed = distance / time
40 = distance / 3
Cross multiply
Distance = 40 × 3
Distance = 120 Km
For car B:
Speed = 50 km/h
Time = 3 hours
Distance =?
Speed = distance / time
50 = distance / 3
Cross multiply
Distance = 50 × 3
Distance = 150 Km
Finally, we shall determine the distance between car B an car A. This can be obtained as follow:
Distance travelled by car B (D₆) = 150 Km
Distance travelled by car A (Dₐ) = 120 Km
Distance apart =?
Distance apart = D₆ – Dₐ
Distance apart = 150 – 120
Distance apart = 30 Km
Therefore, car B will be 30 Km ahead of car A.
Answer:
a. The angular frequency is doubled.
e. The period is reduced to one-half of what it was.
Explanation:
Angular frequency is given as;
ω = 2πf
when the frequency is doubled
Thus, the angular frequency will be doubled.
Amplitude in simple harmonic motion is the maximum displacement.
Frequency is related to period in simple harmonic motion as given in the equation below;
when the frequency is doubled;
Thus, the period will be reduced to one-half of what it was.