Answer:
i took g = 9.8m/s
A. 1.16secs
B. 2.32secs
C. 6.57m
D. 57.91m
Explanation:
A. How long does the Missile take to reach ot peak?
Time taken (t) =( U²Sin (angle) )/g
u = initial velocity = 25m/s
angle given = 30°
g = acceleration due to gravity = 9.8m/s²
t = U² x Sin (angle) / g
t = 25² x Sin(30)/9.8
t = 1.61secs
B. How long is the missile in the air in total?
T = 2t
T = 2 x 1.61 = 2.32 secs
C. what maximum Height does the missile reach?
- Maximum height = U²Sin²(angle) / 2g
- M.H =25² x Sin(30)² / 2 x 9.8
- M.H=6.57m
- Maximum height= 6.57m
D. How far does the missile travel Horizontally?
- Range = U²2Sin(angle)/g
- Range = 25² x 2 x Sin(30) / 9.8
- Range = 57.91m
<em>To understand the passage between two blades, it is required </em>
<em>to travel the distance of the circumference equivalent to the </em>
<em>segment of the diameter that exists between them,</em>
Where
Ball diameter
Space time
So the angle swept out by either a blade or a space is:
Through the angular velocity
So,
Answer:
A first-class lever: fulcrum is between input and output force; second-class lever: output force is between input force and fulcrum; third-class lever: input force is between fulcrum and output force
Answer:
Explanation:
Let
h = height of balloon (in feet).
θ = angle made with line of sight and ground (in radians).
h = 300 tanθ
now can be written as
When θ = π/4,