A. Direction
Speed is just distance divided by time, but velocity is displacement divided by time and displacement has direction. Speed will always be positive, but velocity can be either positive or negative.
B: asking questions
(Hope this helps)
The law applied here is Hooke's Law which describes the force exerted by the spring with a given distance. The equation for this is F = kΔx, where F is the force in Newtons, k is the spring constant in N/m while Δx is the displacement in meters.
If you want to find work done by a spring, this can be solved by using differential equations. However, derived equations are already ready for use. The equation is
W = k[{x₂-x₁)² - (x₁-xn)²],
where
xn is the natural length
x₁ is the stretched length
x₂ is also the stretched length when stretched even further than x₁
In this case xn =x₁. So, that means that (x₁-xn) = 0 and (x₂-x₁) = 11 cm or 0.11 m.
Then, substituting the values,
2 J = k (0.11² -0²)
k = 165.29 N/m
Finally, we use the value of k to the Hooke's Law to determine the Force.
F = kΔx = (165.29 N/m)(0.11 m)
F = 18.18 Newtons
This is correct. The lighter truck moves faster.
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)