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mario62 [17]
3 years ago
5

At an equilibrium position of a pendulum, the is at a maximum.

Physics
2 answers:
ANEK [815]3 years ago
6 0
<span>The equilibrium position is that at which the pendulum is at its lowest point; it is called this because, absent any other forces acting upon it, this is the point at which the pendulum would be at a stable, motionless equilibrium. It is also the point at which the pendulum, having been released from above, has translated its starting gravitational potential energy fully into kinetic energy. As such, this means that at this point the pendulum is at its maximum D) velocity.</span>
noname [10]3 years ago
5 0
<span>The equilibrium position is the point below the pivot, at a neutral position. amplitude of the pendulum’s swing is the displacement from the equilibrium. each swing is referred to as maximum displacement or maximum amplitude. 

So your answer is D. Velocity</span>
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jolli1 [7]

Answer:

a)V=\dfrac{5.3}{P}

b)ML^{-4}T^{-2}.

Explanation:

Given that

Boyle's law

P V = Constant ,at constant temperature

a)

Given that

P_1=50KPa

V_1=0.106m^3

We know that for PV=C

P_1V_1=P_2V_2=PV

Now by putting the values

PV= 50 x 0.106

V=\dfrac{5.3}{P}

Where P is in KPa and V is in m^3

b)

PV= C

Take ln both sides

So \ln(PV)=\ln C

lnP + lnV =lnC               ( C is constant)

By differentiating

\dfrac{dP}{P}+\dfrac{dV}{V}=0

So

\dfrac{dP}{dV}=-\dfrac{P}{V}

When P= 50 KPa

\dfrac{dP}{dV}=-\dfrac{50}{V}\ \dfrac{KPa}{m^3}

It indicates the slope of PV=C curve.

It unit is \dfrac{Pa}{m^3}.

Or we can say that ML^{-4}T^{-2}.

5 0
3 years ago
Read 2 more answers
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