Answer:
so its easier to understand for the reader
Explanation:
Elements with atomic numbers from 58 through 71 are part of the
<span>
lanthanide</span> series <span />
The change in potential energy when the block falls to ground is -480J.
The maximum change in kinetic energy of the ball is 480 J.
The initial kinetic energy of the ball is 0 J.
The final kinetic energy of the ball is 0.148J.
The initial potential energy of the ball is 0.187 J.
The final potential energy of the ball is 0 J.
The work done by the air resistance is 0.039 J.
<h3>Change in potential energy when the block falls to ground</h3>
ΔP.E = -mgh
ΔP.E = -Wh
ΔP.E = - 40 x 12
ΔP.E = -480 J
<h3>Maximum change in kinetic energy of the ball</h3>
ΔK.E = - ΔP.E
ΔK.E = - (-480 J)
ΔK.E = 480 J
<h3>Initial kinetic energy of the ball</h3>
K.Ei = 0.5mv²
where;
- v is zero since it is initially at rest
K.Ei = 0.5m(0) = 0
<h3>Final kinetic energy</h3>
K.Ef = 0.5mv²
K.Ef = 0.5(0.0091)(5.7)²
K.Ef = 0.148 J
<h3>Initial potential energy of the ball</h3>
P.Ei = mghi
P.Ei = 0.0091 x 9.8 x 2.1
P.Ei = 0.187 J
<h3>Final potential energy</h3>
P.Ef = mghf
P.Ef = 0.0091 x 9.8 x 0
P.Ef = 0
<h3>Work done by the air resistance</h3>
W = ΔE
W = P.E - K.E
W = 0.187 J - 0.148 J
W = 0.039 J
Learn more about potential energy here: brainly.com/question/1242059
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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 
Answer:
Current = dQ/dt
or I = dQ/dt
Where I represents current.
Which is the rate of flow of charge.
Q=4 + 2t + t²
dQ/dt = 2 + 2t --- This is the relation that gives the instantaneous current.
At time t=2sec
dQ/dt = I = 2 + 2t
= 2 + 2(2)
=2 + 4
= 6A.