1.0 joule= 1.0 newtons × 1.0 meter = 1.0 newton × meter
Work = 10 newtons × 5 meters = 50 newton × meter
The easiest way to answer this question is by realizing there are relating the velocities of the two cars. To tackle this problem, you have to understand the picture. Car 1 travels at 35m/s and Car 2 travels at 25m/s. Based on relative velocities, we can understand that Car 1 travels 10m/s faster than Car 2 every second. So we can interpret Car 1's relative velocity to Car 2 as 10m/s. Car 1 needs to travel 10m/s till a point of catching up to Car 2 which is 462m away.
v = 10m/s
d = 462m
v = d/t
(10) = (462)/t
t = 46.2s
So it takes 46.2 seconds for Car 1 to catch up to Car 2, but the question is asking how far does Car 1 travel to catch up. So we have to use Car 1's velocity and not the relative velocity:
v = 35m/s
v = d/t
(35) = d/(46.2)
d = 1617m
Car 1 traveled a total distance of 1617m.
Answer:
vi = 3.95 m/s
Explanation:
We can apply the Work-Energy Theorem as follows:
W = ΔE = Ef - Ei
W = - Ff*d
then
Ef - Ei = - Ff*d <em> </em>
If
Ei = Ki + Ui = 0.5*m*vi² + m*g*hi = 0.5*m*vi² + m*g*hi = m*(0.5*vi² + g*hi)
hi = d*Sin 20º = 5.1 m * Sin 20º = 1.7443 m
Ef = Kf + Uf = 0 + 0 = 0
As we know, vf = 0 ⇒ Kf = 0
Uf = 0 since hf = 0
we get
W = ΔE = Ef - Ei = 0 - m*(0.5*vi² + g*hi) ⇒ W = - m*(0.5*vi² + g*hi) <em> (I)</em>
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If
W = - Ff*d = - μ*N*d = - μ*(m*g*Cos 20º)*d = - μ*m*g*Cos 20º*d <em>(II)</em>
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we can say that
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- m*(0.5*vi² + g*hi) = - μ*m*g*Cos 20º*d
⇒ vi = √(2*g*(μ*Cos 20º*d - hi))
⇒ vi = √(2*(9.81 m/s2)*(0.53*Cos 20º*5.1m - 1.7443 m)) = 3.95 m/s
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The most waste is Yucca mtn. in Nevada
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If an ion is negatively charged, it has gained electrons and now has more electrons than protons because before the atom became ionized it was neutrally charged.
