Answer:
m₁ (R +ℓ/2) / /( m₁ + m₂ )
Explanation:
We shall consider centre of the club's head as origin .
The centre of mass of the rod attached with sphere will be away from the centre of sphere by a distance
= R + ℓ /2 and its mass is m₁
mass of sphere is m₂ and distance of its centre of mass from origin is zero
So required CM = m₁ x₁ + m₂ x₂ / ( m₁ + m₂ )
= m₁ (R + ℓ /2) + m₂ x 0 /( m₁ + m₂ )
= m₁ (R +ℓ/2) / /( m₁ + m₂ )
speed of the plane is given as

height of the plane is

acceleration due to gravity is

part a)
when package is dropped its vertical speed is zero so we can use kinematics to find the time of drop



Part b)
Horizontal distance moved by the package is given by



Part c)
final speed in x direction

final speed in y direction

so net speed as it hit the ground will be



Input= ?
Output= 0.75
MA= 0.33
So I just kept on Dividing numbers until I was close to "0.33" & I cam up with the answer of .......... 0.25.
So when you divide 0.25 by 0.75 you get the MA of 0.33
Answer:
t=0.016s
Explanation:
we can use

the time that the particle is in the magnetic field is one half oa period. Hence

I hope this is useful for you
regards
Answer:
The answer would be Saturn.
Explanation:
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