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)
Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.
Note that
1 m = 1609.34 m.
Therefore
v = (60 mi/h)*(1609.34 m/mi)*(1/3600 h/s) = 26.822 m/s
Use the formula
v = u + at
(26.822 m/s) = (2.35 m/s²)*(t s)
t = 26.822/2.35 = 11.4 s
Answer: 11.4 s
Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
Therefore
(26.822 m/s) = (a m/s²)*(0.6 s)
a = 26.822/0.6 = 44.7 m/s²
Answer: 44.7 m/s²
Take into account that in a standing wave, the frequency f of the points executing simple harmonic motion, is simply a multiple of the fundamental harmonic fo, that is:
f = n·fo
where n is an integer and fo is the first harmonic or fundamental.
fo is given by the length L of a string, in the following way:
fo = v/λ = v/(L/2) = 2v/L
becasue in the fundamental harmonic, the length of th string coincides with one hal of the wavelength of the wave.
The formula that can be used to show that the speed of the stone when it hits the ground is 32 m/s is v = u + at
According to equations of motion,
v = u + at
v = Final velocity
u = Initial velocity
t = Time
a = Acceleration due to gravity
t = 3.2 s
a = 10 m / s²
v = 0 + ( 10 * 3.2 )
v = 32 m /s
Equations of motion are used to describe the position of a moving object in a constantly accelerated manner. The position can be described using acceleration, velocity, displacement and time.
Therefore, the formula that can be used to show that the speed of the stone when it hits the ground is 32 m/s is v = u + at
To know more about equations of motion
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