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Rudiy27
3 years ago
11

A slender uniform rod 100.00 cm long is used as a meter stick. two parallel axes that are perpendicular to the rod are considere

d. the first axis passes through the 50-cm mark and the second axis passes through the 30-cm mark. what is the ratio of the moment of inertia through the second axis to the moment of inertia through the first axis?

Chemistry
2 answers:
Stels [109]3 years ago
8 0

Answer:  L2/L1=1.5

Explanation:

Not my work but heres the way the answer is obtained cause the other guy was totally incorrect

Write the expression to calculate the moment of inertia of the meter stick about second axis of rotation. I 2 = M L 2 12 + M ( x 1 − x 2 ) 2

Substitute the values in the above expression. I 2 = M ( 1 m ) 2 12 + M ( 0.5 m − 0.3 m ) 2 I 2 = M 12 + 0.04 M I 2 = 1.48 M 12

. . . . . . ( i i ) Divide expression (ii) by expression (i):

I 2 I 1 = 1.48 M 12 M 12 = 1.48 ≃ 1.5

kvv77 [185]3 years ago
6 0
Refer to the diagram shown below.

The second axis is at the centroid of the rod.
The length of the rod is L = 100 cm = 1 m

The first axis is located at 20 cm = 0.2 m from the centroid.
Let m =  the mass of the rod.

The moment of inertia about the centroid (the 2nd axis) is
I_{g} =  \frac{mL^{2}}{12} = (m \, kg) \frac{(1 \, m)^{2}}{12} =  \frac{m}{12} \, kg-m^{2}

According to the parallel axis theorem, the moment of inertia about the first axis is
I_{1} = I_{g} + (m \, kg)(0.2 m)^{2} \\ I_{1} =  \frac{m}{12}+ 0.04m = 0.1233m \, kg-m^{2}

The ratio of the moment of inertia through the 2nd axis (centroid) to that through the 1st axis is
\frac{I_{g}}{I_{1}} =  \frac{0.0833m}{0.1233m} =0.6756

Answer:  0.676

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Shalnov [3]

Explanation:

Formula to calculate hybridization is as follows.

                Hybridization = \frac{1}{2}[V+N-C+A]

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

So, hybridization of BeCl_{2} is as follows.

              Hybridization = \frac{1}{2}[V + N - C + A]

                                    = \frac{1}{2}[2 + 2]

                                    = 2

Hybridization of BeCl_{2} is sp. Therefore, BeCl_{2} is a linear molecule. There will be only two electron groups through which Be is attached.

Similarly, hybridization of XeF_{2} is calculated as follows.

         Hybridization = \frac{1}{2}[V + N - C + A]

                                    = \frac{1}{2}[8 + 2]

                                    = 5

Therefore, hybridization of XeF_{2} is sp^{3}d. Therefore, [tex]XeF_{2} is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.

Thus, we can conclude that out of the given options XeF_{2} is the correct examples of linear molecules for five electron groups.

7 0
3 years ago
Calculate δ h for the reaction:no (g) + o2 (g) ↔ no2 (g). given: 2o3(g) ↔ 3o2(g) δh=-426 kj o2(g) ↔ 2o(g) δh=+ 490 kj no(g) + o3
maks197457 [2]
To calculate the <span>δ h, we must balance first the reaction: 

NO + 0.5O2 -----> NO2

Then we write all the reactions,

2O3 -----> 3O2    </span><span>δ h = -426 kj        eq. (1)

O2 -----> 2O    </span><span>δ h = 490 kj             eq. (2)

NO + O3 -----> NO2 + O2    </span><span>δ h = -200 kj          eq. (3)


We divide eq. (1) by 2, we get

</span>O3 -----> 1.5O2    δ h = -213  kj             eq. (4)

Then, we subtract eq. (3) by eq. (4) 

NO + O3 ----->  NO2 + O2   δ h = -200 kj
-       (O3 -----> 1.5 O2         δ h = -213  kj)
NO -----> NO2 - 0.5O2        δ h = 13  kj               eq. (5)


eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)

O -----> 0.5O2  <span>δ h = -245  kj         eq. (6)
</span>
Add eq. (6) to eq. (5), we get

NO -----> NO2 - 0.5O2        δ h = 13  kj 
+  O -----> 0.5O2                 δ h = -245  kj
NO + O ----> NO2               δ h = -232 kj

<em>ANSWER:</em> <em>NO + O ----> NO2               δ h = -232 kj</em>


4 0
3 years ago
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n200080 [17]

Answer: has diffused more quickly, so it must be a

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expect a solid to be more dense than its liquid,

and therefore that ice would sink in water.

8 Bubbles will be present through the volume

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speed. The inter-particle distances in the gas are

significantly larger than those in the liquid.

9 At certain conditions of low temperature and low

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phase.

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as it forms liquid at the same temperature (E–D

on Figure 1.4). This is additional to the energy

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Explanation:

4 0
3 years ago
Consider the following weak acids and their Ka values: Acetic Acid Ka = 1.8×10^−5 Phosphoric Acid Ka = 7.5×10^−3 Hypochlorous Ac
vfiekz [6]

Answer:

a. Phosphoric Acid

b. Acetic Acid

c. Hypochlorous Acid

Explanation:

A buffer works when the pH of this one is in pKa ± 1. That means, to find which buffer system works in some pH you need to find pKa:

pKa = -log Ka

<em>pKa Acetic acid:</em>

-log1.8x10⁻⁵ = 4.74

<em>pKa phosphoric acid:</em>

-log7.5x10⁻³ = 2.12

<em>pKa hypochlorous acid:</em>

-log3.5x10⁻⁸ = 7.46

a. For a pH of 2.8 the best choice is phophoric acid because its effective range is: 1.12 - 3.12 and 2.8 is between these values.

b. pH 4.5. Acetic acid. effective between pH's 3.74 - 5.74

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7 0
3 years ago
When CO2(g) is put in a sealed container at 730 K and a pressure of 10.0 atm and is heated to 1420 K , the pressure rises to 24.
d1i1m1o1n [39]

Answer:

48%

Explanation:

Based on Gay-Lussac's law, the pressure is directly proportional to the temperature. To solve this question we must assume the temperature increases and all CO2 remains without reaction. The equation is:

P1T2 = P2T1

<em>Where Pis pressure and T absolute temperature of 1, initial state and 2, final state of the gas:</em>

P1 = 10.0atm

T2 = 1420K

P2 = ?

T1 = 730K

P2 = 10.0atm*1420K / 730K

P2 = 19.45 atm

The CO2 reacts as follows:

2CO2 → 2CO+ O2

Where 2 moles of gas react producing 3 moles of gas

Assuming the 100% of CO2 react, the pressure will be:

19.45atm * (3mol / 2mol) = 29.175atm

As the pressure rises just to 24.1atm the moles that react are:

24.1atm * (2mol / 19.45atm) = 2.48 moles of gas are present

The increase in moles is of 0.48 moles, a 100% express an increase of 1mol. The mole percent that descomposes is:

0.48mol / 1mol * 100 = 48%

8 0
3 years ago
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