2000J
Explanation:
Given parameters:
Extension = 0.5m
Spring constant = 16000N/m
Unknown:
Energy stored in the bow string = ?
Solution:
The energy stored in a bow string is an elastic potential energy.
It can be calculated using the expression below;
Elastic energy = 
K e²
Where k is the spring constant
e is the extension
Input the parameters;
Elastic energy = K e²
= x 16000 x 0.5²
= 2000J
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Answer:
Explanation:
a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force
R = mg where m is mass of the block
Force of friction F = μ x mg
= .173 x 12.2 x 9.8
= 20.68 N
b ) Only force of friction is acting on the body so
deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²
acceleration = - 1.7 m /s²
c )
v² = u² - 2 a s
v = 0 , u = 3.9 m /s
a = 1.7 m /s
0 = 3.9² - 2 x 1.7 x s
s = 4.47 m
Answer:
Time taken, 
Explanation:
It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.
From the figure,
The sum of forces in y direction is :
Sum of forces in x direction,
.............(1)
Also, 
Equation (1) becomes :
...............(2)
Let t is the time taken for the ball to rotate once around the axis. It is given by :
Put the value of T from equation (2) to the above expression:
On solving above equation :
Hence, this is the required solution.
