A string with linear density 0.500 g/m.
Tension 20.0 N.
The maximum speed 
The energy contained in a section of string 3.00 m long as a function of .
We are given following data for string with linear density held under tension :
μ = 0.5 
= 0.5 x 10⁻³ 
T = 20 N
If string is L = 3m long, total energy as a function of is given by:
E = 1/2 x μ x L x ω² x A²
= 1/2 x μ x L x 
= 7.5 x 10⁻⁴
So, The total energy as a function of = 7.5 x 10⁻⁴
Learn more about linear density problem here:
brainly.com/question/17190616
#SPJ4
Question: A man pushes down on a table. what does the table do to the man?
Answer: pushes back against the hand
Explanation: The weight of the ma pushes down on the table with a force mg while the table pushes up on the book with an equal and opposite force
question answered by
(jacemorris04)
<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:
i d k about that but I know it`s a Polish thing
