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LiRa [457]
3 years ago
12

A wheel is rotating freely at angular speed 420 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initi

ally at rest and with 7 times the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels
Physics
1 answer:
Mashcka [7]3 years ago
8 0

Answer:60 rev/min

Explanation:

Given

angular speed of first shaft \omega _1=420\ rev/min

Moment of inertia of second shaft is seven times times the rotational speed of the first i.e. If I is the moment  of inertia of first wheel so moment of inertia of second is 7 I

As there is no external torque therefore angular momentum is conserved

L_1=L_2

I_1\omega _1=I_2\omega _2

I\times (420)=7 I\times (\omega _2)

\omega _2=\frac{420}{7}

\omega _2=60\ rev/min  

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Gary saved dimes and nickels at a ratio of 5:7. Then, he saved 20% more dimes and 3 times the original number of nickels. If he
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Answer:

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Explanation:

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3 years ago
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A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

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A 0.200-kg cube of ice (frozen water) is floating in glycerine.The gylcerine is in a tall cylinder that has inside radius 3.90 c
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Answer:

Part a)

h = 0.86 cm

Part b)

Level will increase

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Part a)

Mass of the ice cube is 0.200 kg

Now from the buoyancy force formula we know that weight of the ice is counter balanced by buoyancy force on the ice

So here we will have

mg = \rho V_{displaced} g

V_{displaced} = \frac{m}{\rho}

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so here volume of water that will convert due to melting of ice is given as

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So here extra volume that rise in the level will be given as

\Dleta V = V - V_{displaced}

\pi r^2 h = 2\times 10^{-4} - 1.59 \times 10^{-4}

(\pi (0.039^2) h = 0.41 \times 10^{-4}

h = 0.86 cm

Part b)

Since volume of water that formed here is more than the volume that is displaced by the ice so we can say that level of liquid in the cylinder will increase due to melting of ice

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