Answer:
h₍₁₎ = 495,1 meters
h₍₂₎ = 480,4 m
h₍₃₎ = 455,9 m
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Explanation:
The exercise is "free fall". t =
Solving with this formula you find the time it takes for the stone to reach the ground (T) = 102,04 s
The heights (h) according to his time (t) are found according to the formula:
h(t) = 500 - 1/2 * g * t²
Remplacing "t" with the desired time.
Answer:1200
Explanation:
Given data
Upper Temprature
Lower Temprature
Engine power ouput
Efficiency of carnot cycle is given by
rounding off to two significant figures
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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