Answer:
Vibrating means to move quickly to and fro.
Explanation:
The acceleration that Andrew experience during his ride is 3.6m/s²
The formula for calculating centripetal acceleration is expressed as:
a = v²/r
v is the speed
r is the radius
Given the following expression
v = 6m/s
r = 10m
Substitute the given parameters into the formula
a = 6²/10
a = 36/10
a = 3.6m/s²
Hence the acceleration that Andrew experience during his ride is 3.6m/s²
Learn more here: brainly.com/question/1268866
Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.
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:
![Vt = \sqrt{(6)^{2} +(-6)^{2} } \\Vt=8.48[km/h]](https://tex.z-dn.net/?f=Vt%20%3D%20%5Csqrt%7B%286%29%5E%7B2%7D%20%2B%28-6%29%5E%7B2%7D%20%7D%20%5C%5CVt%3D8.48%5Bkm%2Fh%5D)
Answer
given,
before collision
mass of car A = m_a = 1300 kg
velocity of car A = v_a = 35 mph
mass of car B = m_b= 1000 kg
velocity of car B = v_b = 25 mph
after collision
V_a = 30 mph
V_b = 31.5 mph
Initial momentum



final momentum



here initial momentum is equal to the final momentum of the car.
hence, momentum is conserved in the collision.