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erastovalidia [21]
4 years ago
6

When visiting some countries, you may see a person balancing a load on the head. Explain why the center of mass of the load need

s to be directly above the person’s neck vertebrae.
Physics
1 answer:
Vinvika [58]4 years ago
3 0

Explanation:

If the center of the load is directly above the vertebrae, there is no torque in the system. This is a good thing so that the vertebrae are not put out of alignment over time. (Of course, this still doesn't prevent compression of the vertebrae over time, which is a possibility.)

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Find velocity vx(t) and coordinate x(t) for a particle of mass m which is subject to the force given by: Fx = F0 e −kt , where F
muminat

Answer:

v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})

x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)

Explanation:

Given that the force of the particle is,

F_{x}=F_{0}e^{-kt}

Now it can be further written as

m\frac{dv}{dt}= F_{0}e^{-kt}\\\frac{dv}{dt}=\frac{ F_{0}}{m} e^{-kt}\\dv=\frac{ F_{0}}{m}e^{-kt}dt\\ v=\frac{ F_{0}}{-km}e^{-kt}+C

Now the initial conditions are v=1 at t=0.

So,

1=\frac{ F_{0}}{-km}e^{0}+C\\C=1+\frac{ F_{0}}{km}

Now the velocity will become.

v_{x} (t)=\frac{ F_{0}}{-km}e^{-kt}+1+\frac{ F_{0}}{km}\\v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})

And,

\frac{dx}{dt} =1+\frac{ F_{0}}{km}(1-e^{-kt})\\dx=(1+\frac{ F_{0}}{km}(1-e^{-kt}))dt\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+C\\

And, another initial condition is x=0 at t=0

0=0+\frac{ F_{0}0}{km}+\frac{ F_{0}t}{k^{2} m}e^{0}+C\\C=-\frac{ F_{0}t}{k^{2} m}

Now,

x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+-\frac{ F_{0}t}{k^{2} m}\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)

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Answer:

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Explanation:

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2 years ago
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ser-zykov [4K]
Balance tubes by spacing them equally around the centrifuge and Always balance tubes with other tubes containing a same volume of liquid are right. 
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3 years ago
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sashaice [31]
A because of the object isn’t rolling yet and it’s circular then it has the potential to roll
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