Answer:
- <u>59.0891 g (rounded to 4 decimal places)</u>
Explanation:
<em>Half-life time</em> of a radioactive substance is the time for half of the substance to decay.
Thus, the amount of the radioactive substance that remains after a number n of half-lives is given by:
Where:
- A is the amount that remains of the substance after n half-lives have elapses, and
- A₀ is the starting amount of the substance.
In this problem, you have that the half-live for your sample (polonium-210) is 138 days and the number of days elapsed is 330 days. Thus, the number of half-lives elapsed is:
- 330 days / 138 days = 2.3913
Therefore, the amount of polonium-210 that will be left in 330 days is:
Answer:
3676.44 rad/min
Step-by-step explanation:
It is a problem about the angular speed of the car's wheel.
You can calculate the angular speed by using the following formula, which relates the tangential speed of the wheels (the same as the speed of the car) with the angular speed:
( 1 )
v: speed of the car = tangential speed of the wheels = 47mph
r: radius of the wheels = 27/2 in = 13.5 in
you change the units of the speed:
next, you replace the values of v and r in the equation (1):
Then, the car's tires are turning with an angular speed of 3676.44 rad/min
AEB = CED = 180 - 45 - 14 = 121 deg
EDC = 180 - 121 - 27 = 32 deg
So angle D is 32 degrees
Answer:
Angle A is the greatest.
Step-by-step explanation:
