It’s D. An enlargement (hope this helps!)
<span>THIS IS A GAS PHASE REACTION AND WE ARE GIVE PARTIAL PRESSURES . I WRITE IN TERMS OF P RATHER THAN CONCENTRATION :
lnPso2cl12=-kt+lnPso2cl1
initial partial pressure Pso2cl12 the rate constant k and the time t
lnPso2cl12=(4.5*10-2*s-1)*65*s+ln (375)
so lnPso2cl12=3.002
we take the base e antilog:
lnPso2cl12=e3.002
Pso2cl12=20 torr
we use the integrated first order rate
lnPso2cl12=3.002=k*t+ lnPso2cl12=3.002
we use the same rate constant and initial pressure
k=4.5*10-2*s-1
Pso2cl12=375
Pso2cl12=1* so2cl12
Pso2cl12=37.5 torr
subtract in Pso2cl12 grom both side
lnPso2cl12- lnPso2cl12=-kt
ln(x)-ln(y)=ln (x/y)
ln (Pso2cl12/Pso2cl20)=-kt
we get t
-1/k*ln(Pso2cl12/Pso2cl20)=t
t=51 s</span>
I think you want to determine the exit speed?
You have to determine how much velocity was decreased by calculating it from the kinetic energy.
KE = (1/2)mv²
1.4 x 10^5 = (1/2)*(1100)v²
v² = 254.55
v =15.95 m/s
So the velocity reduces by 15.95 m/s. Subtracting this to the initial velocity: 22 - 15.95 = 6.05 m/s.
So, the final speed was 6.05 m/s.
I hope I was able to help :)
Answer:
don't know what class are you you are using which mobile or laptop
Answer:
Tension = 60 N
Explanation:
The ball is executing a motion like a pendulum. The lowest point of a pendulum's motion is the mean position, where its potential energy becomes zero and kinetic energy becomes maximum. At this point the rope becomes straight vertical. Hence, the tension in the rope becomes exactly equal to the weight of the object at the lowest point. Also, in this case when the ball swings through the lowest point during its motion like a pendulum, the tension in the rope will become equal to the weight of the ball.
Tension = Weight
<u>Tension = 60 N</u>
