Answer: W = 0.3853 J, e = 0.052 m
Explanation: Given that,
K =285.0N/M , L = 0.230m , F = 15N , e = ?
F = Ke
15 = 285 × e
e = 15÷ 285
e =0.052 m
e + L = 0.052 + 0.230
= 0.282m ( spring new length )
Work needed to stretch the spring
W = 1/2ke2
W = 1/2 × 285 x 0.052 × 0.052
W = 0.3853 J
The complete queston is The amount of a radioactive element A at time t is given by the formula
A(t) = A₀e^kt
Answer: A(t) =N e^( -1.2 X 10^-4t)
Explanation:
Given
Half life = 5730 years.
A(t) =A₀e ^kt
such that
A₀/ 2 =A₀e ^kt
Dividing both sides by A₀
1/2 = e ^kt
1/2 = e ^k(5730)
1/2 = e^5730K
In 1/2 = 5730K
k = 1n1/2 / 5730
k = 1n0.5 / 5730
K= -0.00012 = 1.2 X 10^-4
So that expressing N in terms of t, we have
A(t) =A₀e ^kt
A₀ = N
A(t) =N e^ -1.2 X 10^-4t
Both don’t have units beacuse they are ratios
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
