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Four identical particles of mass 0.980 kg each are placed at the vertices of a 4.14 m x 4.14 m square and held there by four mas

Zielflug [23.3K]

Answer:

a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²

b) I = 50.39 kg m²

c) I = 16.8 kg m²

Explanation:

a) Given data:

m = 0.98 kg

a = 4.14 * 4.14

The moment of inertia is:

$I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}$

For 4 particles:

$I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}$

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

$I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}$

c)

$I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}$

8 0

2 years ago

Which statements correctly compare the gravitational force with the electrical force? Check all that apply.

skad [1K]

The statements that correctly compare the gravitational force with the electrical force are the following:

-The gravitational force can be attractive.

-The electrical force can be repulsive.

-The electrical force can be attractive.

-Any two objects experience a gravitational force between them.

9 0

3 years ago

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Use V=Voe^-t/RC and T=RC to derive T1/2=Tln2

Ksju [112]

Enclosed is some guidance algebra.I find this q a little confusing. It quotes "RC" which usually makes me think of electrical circuits and time constants based on converting calculating RC value and equating that to t for one time constant then 2RC for two time constants etc. The theory being that after 5 time constants - 5RC - a circuit is stable. BUT, this q then goes on to mention HALF LIFE. The curves for both half life and time constant are both exponential, as in the number e to the power of something, but the algebra is slightly different. I hope my algebra is ok.

3 0

3 years ago

A gas, behaving ideally, has a pressure P1 and at a volume V1. The pressure of the gas is changed to P2. Using Avogadro’s, Charl

Bond [772]

Answer:

Boyle's Law

$\therefore P_1.V_1=P_2.V_2$

Explanation:

Given that:

initially:

pressure of gas, $= P_1$

volume of gas, $= V_1$

finally:

pressure of gas, $= P_2$

volume of gas, $= V_2$

To solve for final volume $V_2$

According to Avogadro’s law the volume of an ideal gas is directly proportional to the no. of moles of the gas under a constant temperature and pressure.

According to the Charles' law, at constant pressure the volume of a given mass of an ideal gas is directly proportional to its temperature.

But here we have a change in the pressure of the Gas so we cannot apply Avogadro’s law and Charles' law.

Here nothing is said about the temperature, so we consider the Boyle's Law which states that at constant temperature the volume of a given mass of an ideal gas is inversely proportional to its pressure.

Mathematically:

$P_1\propto \frac{1}{V_1}$

$\Rightarrow P_1.V_1=k\ \rm(constant)$

$\therefore P_1.V_1=P_2.V_2$

5 0

3 years ago

Lena helps a child who is choking on his food. Which reliable method prepared her for helping with this injury?

Ivanshal [37]

knowledge of first aid ... eg St John Ambulance, Red Cross etc. I think that everyone in a school should be taught First Aid.

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2 years ago

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