Complete Question
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?
Answer:


Explanation:
From the question we are told that
Mass of ball 
Length of string 
Wind force 
Generally the equation for
is mathematically given as




Max angle =
Generally the equation for max Height
is mathematically given as



Generally the equation for Equilibrium Height
is mathematically given as



-- find the horizontal and vertical components of F1.
-- find the horizontal and vertical components of F2.
-- find the horizontal and vertical components of F3.
-- add up the 3 horizontal components; their sum is the horizontal component of the resultant.
-- add up the 3 vertical components; their sum is the vertical component of the resultant.
-- the magnitude of the resultant is the square root of (vertical component^2 + horizontal component^2)
-- the direction of the resultant is the angle whose tangent is (vertical component/horizontal component), starting from the positive x-direction.
The correct answer would be B. A virus is living BUT however it does NOT have all of the 7 characteristics of life.
Hope that helped ^^
Answer:
T = 5.36 s
Explanation:
given,
depth of the mine shaft = 122.5 m
speed of the sound = 340 m/s
time taken = ?
time taken by the stone to reach at the bottom
using equation of motion

initial speed , u = 0 m/s


t = 5 s
time taken by the sound to travel
d =v x t


t = 0.36 s
total time taken for the sound to reach carol after dropping the stone
T = 5 + 0.36
T = 5.36 s