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
Answer:
FB = 0.187 N
Explanation:
To find the magnetic force FB in the wire you use the following formula:

the angle between B and L is given by:

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:
![|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N](https://tex.z-dn.net/?f=%7C%5Cvec%7BF_B%7D%7C%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7DB%28y%29dy%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7D%280.5y%29dy%5C%5C%5C%5C%7C%5Cvec%7BF_B%7D%7C%3D%282.0%2A10%5E%7B-3%7DA%29%28sin36.86%5C%C2%B0%29%280.5T%29%5B%5Cfrac%7B0.25%5E2%7D%7B2%7Dm%5D%3D0.187%5C%20N)
hence, the magnitude of the magnetic force is 0.187N
Force=mass*acceleration
F=ma
F=25*5
F=100 N
Answer:
The answer is 12.67 TMU
Explanation:
Recall that,
worker’s eyes travel distance must be = 20 in.
The perpendicular distance from her eyes to the line of travel is =24 in
What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?
Now,
We solve for the given problem.
Eye travel is = 15.2 * T/D
=15.2 * 20 in/24 in
so,
= 12.67 TMU
Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU
Our values can be defined like this,



The problem can be solved for part A, through the Work Theorem that says the following,

Where
KE = Kinetic energy,
Given things like that and replacing we have that the work is given by
W = Fd
and kinetic energy by

So,

Clearing F,

Replacing the values


B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.