Answer: 
Explanation:
If we have an initial amount of a radioactive material or substance, which is 
, and we also are told this amount decays 
each year, this means each year 
of the substance remains:
<u />
<u>To understand it better:</u>
Year 1: 
Year 2: 
Year 3: 
and so on until year 
:
Year t: 
Therefore, the function tha best describes this radiation decay situation is:
Answer:
The answer is A.
Explanation:
The friction generated heat to release chemical energy because the same same thing happens with fluorescent and sodium vapor lamps; heat causes the (low-pressure) mercury/sodium vapor gas lamp to cause a lot of heat energy resulting in the same thing happening when you strike a match.
Let us first calculate heat obtained by the evaporation of 51 g of water.
Given, heat of vaporization of water = 2.4 kJ/ g
∴ Heat obtained by evaporation of 51 g of water = 2.4 × 51 = 122.4 kJ
This is the heat energy available that can be used to cool water from 42°C to 20°C.
Specific heat of water is given by,
Here,
C is the specific heat of water = 4.18 J/gK
Q is the amount of heat = 122400 J
m is the mass of the water that can be cooled.
dt is the change in temperature= 42°C ₋ 20°C = 22°C ( The numerical value will be the same if Kelvin unit is used.)
Substituting the values we get,
m = 1331 g
1331 grams of water can be cooled from 42°C to 20°C by evaporation of 51 g of water.
Yes it does (not to be mean its kinda stupid for you to ask)
Answer:
Explanation:
We shall solve this problem on the basis of averaging , otherwise it will require integration which is a complex operation .
According to newton's law of cooling
dQ / dt = k ( T₁ - T₂ )
T₁ is temperature of surrounding and T₂ is temperature of object .
For the heating by 2 degree
dQ = ms x ΔT , m is mass , s is specific heat and ΔT is rise of
= ms x 2
dt = 1 second
T₁ the average temperature of object = (30 + 32) / 2 = 31
dQ = ms x ΔT
ms x 2 = k ( 100 - 31 )
k = 2 ms / 69
In the second case bar's temperature rises from 32 to 70
average temperature = 32 + 70 / 2 = 51
If t be the time required
dQ = ms x ( 70 - 32 ) = 38ms
38ms / t = k ( 100 - 51 )
38ms / t = (2ms / 69) x 49
t = 38 x 69 / (2 x 49)
= 26.75 s .
