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Lemur [1.5K]
1 year ago
5

In the diagram, disk 1 has a moment of inertia of 3.4 kg · m2 and is rotating in the counterclockwise direction with an angular

velocity of 6.1 rad/s about a frictionless rod passing through its center. A second disk rotating clockwise with an angular velocity of 9.3 rad/s falls from above onto disk 1. The two then rotate as one in the clockwise direction with an angular velocity of 1.8 rad/s. Determine the moment of inertia, in kg · m2, of disk 2.
Physics
1 answer:
nlexa [21]1 year ago
4 0

Answer:

I = 3.6 kg•m²

Explanation:

Conservation of angular momentum

Let's assume CW is the positive direction

3.4(-6.1) + I(9.3) = 3.4(1.8) + I(1.8)

I(9.3 - 1.8) = 3.4(1.8 + 6.1)

I(7.5) = 3.4(7.9)

I = 3.4(7.9)/(7.5) = 3.5813333333...

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Suppose Earth's mass increased but Earth's diame-
navik [9.2K]

Answer: It would increase.

Explanation:

The equation for determining the force of the gravitational pull between any two objects is:

F = G \frac{m1m2}{r^2}

Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.

Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.

Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.

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2 years ago
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te
QveST [7]

Answer:

Please see attachment

Explanation:

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8 0
1 year ago
The total negative charge on the electrons in 1kg of helium (atomic number 2, molar mass 4) is____________.
tekilochka [14]

Answer:

Explanation:

n = \frac{m}{M}

n = \frac{1000}{4}

         = 250 moles.

    N  = n×6.02×10^{23}

        = 1.505×10^{26}

Total charge = (1.505×10^{26}) × (1.6×10^{-19})

                     = 2.4×10^{7} C.

4 0
2 years ago
Read 2 more answers
a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres
shusha [124]

Answer:

Part a) When collision is perfectly inelastic

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b) When collision is perfectly elastic

v_m = \frac{m + M}{2m}\sqrt{5Rg}

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

mv_m = (m + M)v

so we have

v = \frac{mv_m}{m + M}

now we know that in order to complete the circle we will have

v = \sqrt{5Rg}

\frac{mv_m}{m + M} = \sqrt{5Rg}

now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

Now we know that collision is perfectly elastic

so we will have

v = \frac{2mv_m}{m + M}

now we have

\sqrt{5Rg} = \frac{2mv_m}{m + M}

v_m = \frac{m + M}{2m}\sqrt{5Rg}

6 0
2 years ago
The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al
SashulF [63]

Answer:

7.9 \frac{m}{s^{2} }

Explanation:

Take the fact that mass is inversely proportional to accelertation:

m ∝ a

Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\

Using algebra, we can rearrange our equation to find the final acceleration, a_{2}:

a_{2}  = \frac{a_{1}*m_{1}  }{m_{2} } \\

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:

0.85g * 9.8 \frac{m }{s^{2} } = 8.33 \frac{m }{s^{2} }

Plug everything in:

7.9 \frac{m }{s^{2} } = \frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}

(1590kg the initial weight plus the weight of the added passenger)

8 0
1 year ago
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