A 432 g sample of 60/27Co has a decay constant of 4.14 x 10-9 s-1. How long will it take before only 1/3 of the original sample
1 answer:
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Answer:
1.
109.6 cm , - 1.74 , real
2.
1.5
Explanation:
1.
d₀ = object distance = 63 cm
f = focal length of the lens = 40 cm
d = image distance = ?
using the lens equation
d = 109.6 cm
magnification is given as
m = - 1.74
The image is real
2
d₀ = object distance = a
d = image distance = - (a + 5)
f = focal length of lens = 30 cm
using the lens equation
a = 10
magnification is given as
m = 1.5
Answer:
(a) 1.414 km
(b) 1.06 m/s
Explanation:
(a) For John:
Distance = 1 km north and then 1 km east
Speed = 1.5 m/s
total distance traveled = 1 + 1 = 2 km = 2000 m
Time taken to travel = Distance / speed
t = 2000 / 1.5 = 1333.3 seconds
Displacement =
(b) For jane :
Time is same as john = 1333.33 second
Distance = 1.414 km = 1414 m
Speed = distance / time = 1414 / 1333.33 = 1.06 m/s