Answer:
Explanation:
spring constant k = 425 N/m
a ) At the point of equilibrium
restoring force = frictional force
= kx = 10 N
425 x = 10
x = 2.35 cm
b )
Work done by frictional force
= -10 x 2.35 x 10⁻² x 2 J ( Distance is twice of 2.35 cm )
= - 0.47 J
= Kinetic energy remaining with the cookie as it slides back through the position where the spring is unstretched .
= 425 - 0.47
= 424.53 J
=
<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>
Answer: a system
Explanation: just did the test
Answer:
b. He has to apply less force to lift the box because the mass has decreased.
Explanation: i just took a ck12 test and it was right lol
The equation we want to use is F = ma
mass x acceleration = Force
0.075 kg x 70 m/s = 5.25 Newtons