The shot putter should get out of the way before the ball returns to the launch position.
Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.
The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s
t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.
Answer: 0.45 s
Answer:
Centre of mass of any body is a point where all mass of a body is supposed to be concentrated
it lies in geometrical centre....
Answer:
Explanation:
1 meter = 39.37 inches
meters x inches/meters = inches
6.23x10^-4 m x 39.37 in./m=0.245...=2.45x10^-2
Answer:
Number of electrons are flowing per second is 2.42 x 10¹⁹
Explanation:
The electric current flows through a wire is given by the relation :
....(1)
Here I is current, e is electronic charge, v is drift velocity of electrons and A is the Area of the wire.
But electric current is also define as rate of electrons passing through junction times their charge, i.e. ,
....(2)
Here N is the rate of electrons passing through junction.
From equation (1) and (2).


But area of wire, 
Here d is diameter of wire.
So, 
Substitute 2.91 x 10⁻³ m for d, 0.000191 m/s for v and 6 x 10²⁸ m⁻³ for n in the above equation.

N = 2.42 x 10¹⁹ s⁻¹