The mechanical advantage of a simple machine is 3 and its velocity ratio is 4 find its efficiency
1 answer:
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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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Answer:
The ball would hit the floor approximately after leaving the table.
The ball would travel approximately horizontally after leaving the table.
(Assumption: .)
Explanation:
Let denote the change to the height of the ball. Let
denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let
denote the initial vertical velocity of this ball.
If the air resistance on this ball is indeed negligible:.
The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was .
The height of the table was . Therefore, after hitting the floor, the ball would be
below where it was before leaving the table. Hence,
.
The equation becomes:
.
Solve for :
.
In other words, it would take approximately for the ball to hit the floor after leaving the table.
Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at ) until the ball hits the floor.
The ball was in the air for approximately and would have travelled approximately
horizontally during the flight.