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The rocket travelled a maximum height at 1.0102 km.
Given,
The acceleration of a rocket (a) = 12 m/s²
The altitude of the rocket (s) = 0.50 km = 0.5×10³m
The maximum height of the rocket (h) = ?
Solution,
A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.
The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²
(a) = Δv/Δt
Where , Δv is change in velocity and Δt is change in time.
The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.
(v)= Δx/Δt
Where,Δx is the change in position and Δt is change in time & v is velocity.
Therefore we know the equation of motion is written as,
v² = u² +2as
Where, v is final velocity , u is initial velocity , a is acceleration and s is altitude of the rocket.
Then putting the value ,
v² = 0 + ( 2× 10 × 0.5×10³)m/s
v² = m/s
v = 100 m/s
Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,
v² = u² - 2(g)h
h = (v²- u² ) / 2g
h = 10,000/2×9.8
h = 510.2 m
So that the rocket travelled the maximum height ,
(h)= (0.5 km + 510.2m)
(h) = 1.0102 km
Hence, the rocket travelled at the maximum height h is 1.0102 km
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To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.
In this way,
The total mass of the cars would be,
Therefore the friction force at 29Km / h would be,
In this way the tension exerts between first car and locomotive is,
Therefore the tension in the coupling between the car and the locomotive is
Answer:
Explanation:
Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.
Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t
A(0) = 10 g
Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min
The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min
But the concentration is total amount of salt over 350L constant volume
C = A / 350
Therefore our rate of change for salt A' is
A' = 5 - 5A/350 = 5 - A/70
This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is
So
with A(0) = 10
c + 350 = 10
c = 10 - 350 = -340