Plane a travels at 900km/h and plane b travels at 250/5.which plane travels faster
1 answer:
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Answer:
a) α = 1.875
b) t = 8 s
Explanation:
Given:
ω1 = 0
ω2 = 15
theta (angular displacement) = 60 rad
*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*
a) α = ?
(ω2)^2 = (ω1)^2 + 2α(theta)
=
+ 2(α)(60)
225 = 120α
α = 1.875
b)
α = (ω2-ω1)/t
t = (ω2-ω1)/α = (15-0)/1.875 = 8
t = 8 s
Complete Question
A person throws a pumpkin at a horizontal speed of 4.0 m/s off a cliff. The pumpkin travels 9.5m horizontally before it hits the ground. We can ignore air resistance.What is the pumpkin's vertical displacement during the throw? What is the pumpkin's vertical velocity when it hits the ground?
Answer:
The pumpkin's vertical displacement is
The pumpkin's vertical velocity when it hits the ground is
Explanation:
From the question we are told that
The horizontal speed is
The horizontal distance traveled is
The horizontal distance traveled is mathematically represented as
Where t is the time taken
substituting values
=>
Now the vertical displacement is mathematically represented as
now the vertical velocity before the throw is zero
So
Now the final vertical velocity is mathematically represented as
substituting values
1) Initial upward acceleration:
2) Mass of burned fuel:
Explanation:
1)
There are two forces acting on the rocket at the beginning:
- The force of gravity, of magnitude , in the downward direction, where
is the rocket's mass
is the acceleration of gravity
- The thrust of the motor, T, in the upward direction, of magnitude
According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:
(1)
where a is the acceleration of the rocket.
Solving for a, we find the initial acceleration:
2)
When the rocket reaches an altitude of 5000 m, its acceleration has increased to
The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.
We can therefore rewrite eq.(1) as
where
is the new mass of the rocket
Re-arranging the equation and solving for m', we find
And since the initial mass of the rocket was
This means that the mass of fuel burned is