How many times did the original sample lose 50% of its radioactivity ?
-- Start with. . . . . . . . . . . . 12 grams.
-- Lose half of it once. . . . . . 6 grams left.
-- Lose half of it again . . . . . 3 grams left.
-- Lose half of it again . . . . . 1.5 grams left.
-- Lose half of it again . . . . . 0.75 gram left.
-- How many times did it lose half ? 4 times.
-- How long does it take to lose half ? 4.5 days.
(That's why it's called the 'half-life'.)
-- How long did it take to lose half, 4 times ?
(4 x 4.5 days) = 18 days .
Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
<u>ΔS = 49.13 ft</u>
Answer:
Both Thomson and Rutherford used charged particles in their experiments.
Explanation:
Answer:
The soda is being sucket out at a rate of 3.14 cubic inches/second.
Explanation:
R= 2in
S= π*R²= 12.56 inch²
rate= 0.25 in/sec
rate of soda sucked out= rate* S
rate of soda sucked out= 3.14 inch³/sec
