Answer:
0.1066 hours
Explanation:
A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.
t1/2 = ln2/k
t1/2 = ln2/6.5 h⁻¹
t1/2 = 0.1066 h
The half-life of the pesticide is 0.1066 hours.
As we know that
P.E. = mgh
where,
P.E. = Potential energy of the object =?
m= mass of object= 3kg
g= acceleration due to gravity = 9.8 ms^-2
h = height between object and animal = 0 m
Then
P.E. = 3× 9.8 × 0 = 0 Joules or 0J
<em>Have a luvely day!</em>
Should be :
Lead Sulfate Tetrahydrate
