Answer:
= +3,394 103 m / s
Explanation:
We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.
The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s
The moment before the explosion
p₀ = (m₁ + m₂) v₀
After the explosion
pf = m₁ + m₂ 
p₀ = [texpv_{f}[/tex]
(m₁ + m₂) v₀ = m₁ + m₂
Let's calculate the final speed (v1f) of the first stage
= ((m₁ + m₂) v₀ - m₂ ) / m₁
= ((2100 +1160) 4300 - 1160 5940) / 2100
= (14,018 10 6 - 6,890 106) / 2100
= 7,128 106/2100
= +3,394 103 m / s
come the same direction of the final stage, but more slowly
Answer:
M g H / 2 = M g L / 2 initial potential energy of rod
I ω^2 / 2 = 1/3 M L^2 * ω^2 / 2 kinetic energy attained by rod
M g L / 2 = 1/3 M L^2 * ω^2 / 2
g = 3 L ω^2
ω = (g / (3 L))^1/2
Work done = force * distance moved (in direction of the force)
force= mass* acceleration
force=58.1N
58.1*(5.8*10^4)
=3,369,800 J
Answer:
a_total = 2 √ (α² + w⁴)
, a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 + 
²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m
Answer:
25032.47 W
Explanation:
Power is the time rate of doing work, hence,
P = Work done(non conservative) / time
Work done (non conservative) is given as:
W = total K. E. + total P. E.
Total K. E. = 0.5mv²- 0.5mu²
Where v (final velocity) = 7.0m/s, u (initial velocity) = 0m/s
Total P. E. = mgh(f) - mgh(i)
Where h(f) (final height) = 7.2m, h(i) (initial height) = 0 m
=> W = 0.5mv² - mgh(f)
P = [0.5mv² - mgh(f)] / t
P = [(0.5*790*7²) - (790*9.8*7.2)] / 3
P = (19355 + 55742.4) / 3 = 75097.4/3
P = 25032.47 W
