The half life of Carbon-14 is 5730 years, how many years would it take for 7/8 of the original amount to decay?
<span>Can somebody please help with this problem. I *think* I understand the basics of what a half life is. If I learned correctly, its the amount it takes for half of a sample to decay. It should also happen exponentially, 1/2 remaining after one half life, 1/4 after the second, 1/16 after the third etc. I'm still a little shaky though. Could somebody please clarify what exactly a half life is and how it can be determined (i.e. how to find the time it would take for 7/8 to decay) </span>
Answer:
29.4m/s
Explanation:
Given parameters:
Time = 3s
Unknown:
Average velocity = ?
Solution:
To solve this problem, we use the expression below:
v = u + gt
v is the average velocity
u is the initial velocity = 0m/s
g is the acceleration due to gravity = 9.8m/s²
t is the time
So;
v = 0 + (9.8 x 3) = 29.4m/s
Answer:
57 N
Explanation:
Were are told that the force
of gravity on Tomas is 57 N.
And it acts at an inclined angle of 65°
Thus;
The vertical component of the velocity is; F_y = 57 sin 65
While the horizontal component is;
F_x = 57 cos 65
Thus;
F_y = 51.66 N
F_x = 24.09 N
The net force will be;
F_net = √((F_y)² + (F_x)²)
F_net = √(51.66² + 24.09²)
F_net = √3249.0837
F_net = 57 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
