The complete queston is The amount of a radioactive element A at time t is given by the formula
A(t) = A₀e^kt
Answer: A(t) =N e^( -1.2 X 10^-4t)
Explanation:
Given
Half life = 5730 years.
A(t) =A₀e ^kt
such that
A₀/ 2 =A₀e ^kt
Dividing both sides by A₀
1/2 = e ^kt
1/2 = e ^k(5730)
1/2 = e^5730K
In 1/2 = 5730K
k = 1n1/2 / 5730
k = 1n0.5 / 5730
K= -0.00012 = 1.2 X 10^-4
So that expressing N in terms of t, we have
A(t) =A₀e ^kt
A₀ = N
A(t) =N e^ -1.2 X 10^-4t
Hello There!
There are 5,280 feet in 1 mile.
Have A Great Day!
Answer:
Explanation:
Using the principle of moment, assuming the rod is uniform rod of mass 1 kg
the center of mass of the rod will be at 1 m
assuming the system is in equilibrium,
clockwise moment = anticlockwise moment
let the distance of the man shoulder be x from the center of gravity and also is the pivot point
total mass of bucket + mass of honey = 2kg + 3 kg = 5 kg for rear bucket and
2kg + 5 kg = 7 kg for front bucket
( 5kg × ( 1+x)) + ( 1 kg × x) = 7 kg × ( 1 - x)
5 + 5 x + x = 7 - 7x
5 + 6x = 7 - 7x
6x + 7x = 7 - 5
13x = 2
x = 2 / 13 = 0.154 m
the honeybucket man's shoulder is 0.154 m from the center of the pole ( forward ).
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Mechanical Advantage = Force by Hammer / Force by Nail = 160/40 = 4
Answer : The heat change of the cold water in Joules is, 
Explanation :
First we have to calculate the mass of cold water.
As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.
Now we have to calculate the heat change of cold water.
Formula used :
where,
Q = heat change of cold water = ?
m = mass of cold water = 45 g
c = specific heat of water = 
= initial temperature of cold water = 
= final temperature = 
Now put all the given value in the above formula, we get:
Therefore, the heat change of cold water is
