Answer: hello some of your values are wrongly written hence I will resolve your question using the right values
answer:
stiffness = 1.09 * 10^-6 N/m
Explanation:
Given data:
Length ( l ) = 16 m
radius of wire ( r ) = 3.5 m
mass ( m ) = 5kg
<u>Distance stretched ( Δl ) = 4 * 10^-3 m </u> ( right value )
<u>average bond length ( between atoms ) = 2.3 * 10^-10 m </u>( right value)
first step : calculate the area
area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2
γ = MgL / A Δl
= [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]
= 784.8 / 0.165 = 4756.36 N/m^2
hence : stiffness = γ * bond length
= 4756.36 * 2.3 * 10^-10 = 1.09 * 10^-6 N/m
B. all of nature follows uniform, unchaining laws.
Answer:
7557.875 J
Explanation:
Assumimg we need to find work done by the bullet on the gas
The work is found by integrating force over distance
W= 
For the force in this problem we have
= ![[14000x+10000x^2-21000x^3]_{0}^{0.65}](https://tex.z-dn.net/?f=%5B14000x%2B10000x%5E2-21000x%5E3%5D_%7B0%7D%5E%7B0.65%7D)
=7557.875 J
Answer:
19.8 J
Explanation:
According to the law of conservation of energy, the total mechanical energy of the spring (sum of kinetic energy and elastic potential energy) must be conserved:
(1)
where we have
is the initial kinetic energy of the spring, which is zero because the spring starts from rest (2)
is the elastic potential energy of the spring when it is fully stretched
is the kinetic energy of the spring when it reaches the natural length
is the elastic potential energy of the spring when it reaches its natural length, which is zero because the stretch in this case is zero (3)
So
where
k = 440 N/m is the spring constant
is the initial stretching of the spring
Substituting,
And so using eq.(1) and keeping in mind (2) and (3) we find
