Answer: 0.333 h
Explanation:
This problem can be solved using the <u>Radioactive Half Life Formula</u>:
(1)
Where:
is the final amount of the material
is the initial amount of the material
is the time elapsed
is the half life of the material (the quantity we are asked to find)
Knowing this, let's substitute the values and find
from (1):
(2)
(3)
Applying natural logarithm in both sides:
(4)
(5)
Clearing
:
(6)
Finally:
This is the half-life of the Bismuth-218 isotope
Answer:
0.3659
Explanation:
The power (p) is given as:
P = AeσT⁴
where,
A =Area
e = transmittivity
σ = Stefan-boltzmann constant
T = Temperature
since both the bulbs radiate same power
P₁ = P₂
Where, 1 denotes the bulb 1
2 denotes the bulb 2
thus,
A₁e₁σT₁⁴ = A₂e₂σT₂⁴
Now e₁=e₂
⇒A₁T₁⁴ = A₂T₂⁴
or

substituting the values in the above question we get

or
=0.3659
Answer:
B, C, F
Explanation:
B: Sugar can be separated from the water by evaporating the water. This will leave large chunks of sugar.
C: Sugar gets spread out among the water.
F: Sugar water is a homogeneous <u>mixture. </u>Can't see the individual components because of the dissolving.
Hoped this helped! :)
Answer:
Refractive index of a medium = Speed of light in vacuum/Speed of light in a medium
1.5 = 3 x 108 / Speed of light in medium
Speed of light in the medium = 3 x 108 /1.5
= 2 x 108 m/s.
Explanation: