<u>Answer:</u>
Work input = Work output * Work against friction is your answer so C
<u>Explanation:</u>
I hope this helps you :)
If it is s-t graph , point is c
if it is v-t graph , point is e
Answer:
so rate constant is 4.00 x 10^-4 
Explanation:
Given data
first-order reactions
85% of a sample
changes to propene t = 79.0 min
to find out
rate constant
solution
we know that
first order reaction are
ln [A]/[A]0 = -kt
here [A]0 = 1 and (85%) = 0.85 has change to propene
so that [A] = 1 - 0.85 = 0.15.
that why
[A] / [A]0= 0.15 / 1
[A] / [A]0 = 0.15
here t = (79) × (60s/min) = 4740 s
so
k = - {ln[A]/[A]0} / t
k = -ln 0.15 / 4740
k = 4.00 x 10^-4 
so rate constant is 4.00 x 10^-4 
Given the velocity-time graph of an object.
The slope of a velocity-time graph gives the acceleration acting on the object.
From the graph, we can see that the slope of the graph is zero. That is, the velocity of the object is constant and hence the net acceleration acting on the object is zero.
From Newton's second law, the net force acting on an object is given by the product of the mass of the object and its velocity. Therefore when the acceleration of the object is zero, the net force on the object is also zero.
Therefore the net force acting on the given object is zero. Hence, the correct answer is option A.