<span>120 revolutions per min is (20RPM)
120revolutions is 240 Pi radians (because 1 revolution is 2Pi rad)
angular velocity
w= 240Pixf, f is the frequency and f= 1/T, T =1mn=60s=period, so f=1/60=0.01Hz
so w = 240*Pi*0.01=12.56 rad /s
linear velocity can be found with
V =Rx w,
V=12.56 R the value depends on what value is the radius of both wheels
</span>
Answer:
1.25 m
0.5 s
Explanation:
Given:
v₀ = 5 m/s
v = 0 m/s
a = -10 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (5 m/s)² + 2 (-10 m/s²) Δy
Δy = 1.25 m
Find: t
v = at + v₀
(0 m/s) = (-10 m/s²) t + (5 m/s)
t = 0.5 s
<span> the </span>law of conservation of energy<span> states that the total </span>energy<span> of an isolated system remains constant it would be said to be conserved over time. </span>Energy<span> can neither be created nor destroyed but then rather, it transforms from one form to another.
hope this can help you ^_^</span>
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>
<em />
If
W = - Ff*d = - μ*N*d = - μ*(m*g*Cos 20º)*d = - μ*m*g*Cos 20º*d <em>(II)</em>
<em />
we can say that
<em />
- 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
<em />
