Average speed of the driver is given as
if he moved for total time t = 3.5 hours
so the distance between the tow is given as
so the distance between St. Louis and Chicago is 224 miles
Zoe went on a fishing trip. On the first day she caught 9 catfish and 6 trout. On second caught 17 catfish and 13 trought. On wh
1 answer:
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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:
a) t = 20 [s]
b) Can't land
Explanation:
To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.
a)
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = desacceleration = 5 [m/s^2]
t = time [s]
Note: the negative sign of the equation means that the aircraft slows down as it stops.
0 = 100 - 5*t
5*t = 100
t = 20 [s]
b)
Now we can find the distance using the following kinematics equation.
x - xo = distance [m]
x -xo = (0*20) + (0.5*5*20^2)
x - xo = 1000 [m]
1000 [m] = 1 [km]
And the runaway is 0.8 [km], therefore the jetplane needs 1 [km] to land. So the jetpalne can't land