Question 5
1 answer:
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Answer:
a
The radial acceleration is
b
The horizontal Tension is
The vertical Tension is
Explanation:
The diagram illustrating this is shown on the first uploaded
From the question we are told that
The length of the string is
The mass of the bob is
The angle made by the string is
The centripetal force acting on the bob is mathematically represented as
Now From the diagram we see that this force is equivalent to
where T is the tension on the rope and v is the linear velocity
So
Now the downward normal force acting on the bob is mathematically represented as
So
=>
=>
The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as
=>
substituting values
The horizontal component is mathematically represented as
substituting value
The vertical component of tension is
substituting value
The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is
substituting value
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
To know more about acceleration
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