1) The half-life is the time required for a substance to reduce to half its initial value. In formulas:
(1)
where
m(t) is the amount of substance left at time t
m0 is the initial mass
is the half-life
In this problem, the half-life of the substance is 20 years:
therefore, the fraction of sample left after t=40 years will be
So, only 1/4 of the original sample will be left, which corresponds to 25%.
2) We can use again formula (1), by re-arranging it:
If we use m(t)=10 g (mass of uranium left at time t), and
(the time is equal to 4 half lifes), we get
So, the initial sample of uranium was 160 g.
Answer:
Gamma radiation or Cathode rays
Explanation:
by striking incident gamma or cathode rays onto the solid when placed on a photographic plate
Thermal energy moves between objects through conduction. This is because, collisions made from the objects through the molecules.
Heat can also move in three different ways. Convection, conduction, and radiation.
'H' = height at any time
'T' = time after both actions
'G' = acceleration of gravity
'S' = speed at the beginning of time
Let's call 'up' the positive direction.
Let's assume that the tossed stone is tossed from the ground, not from the tower.
For the stone dropped from the 50m tower:
H = +50 - (1/2) G T²
For the stone tossed upward from the ground:
H = +20T - (1/2) G T²
When the stones' paths cross, their <em>H</em>eights are equal.
50 - (1/2) G T² = 20T - (1/2) G T²
Wow ! Look at that ! Add (1/2) G T² to each side of that equation,
and all we have left is:
50 = 20T Isn't that incredible ? ! ?
Divide each side by 20 :
<u>2.5 = T</u>
The stones meet in the air 2.5 seconds after the drop/toss.
I want to see something:
What is their height, and what is the tossed stone doing, when they meet ?
Their height is +50 - (1/2) G T² = 19.375 meters
The speed of the tossed stone is +20 - (1/2) G T = +7.75 m/s ... still moving up.
I wanted to see whether the tossed stone had reached the peak of the toss,
and was falling when the dropped stone overtook it. The answer is no ... the
dropped stone was still moving up at 7.75 m/s when it met the dropped one.
Answer:
32s
Explanation:
We must establish that by the time the police car catches up to the speeder, both have travelled a certain distance during the same amount of time. However, the police car experiences accelerated motion whereas the speeder travels at a constant velocity. Therefore we will establish two formulas for distance starting with the speeder's distance:
and the police car distance:
Since they both travel the same distance x, we can equal both formulas and solve for t:
Two solutions exist to the equation; the first one being
The second solution will be:
This result allows us to confirm that the police car will take 32s to catch up to the speeder