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Jlenok [28]
3 years ago
13

Question 5. Our results support the idea that if left to freely oscillate, a system will vibrate at a natural frequency that dep

ends on the system itself, not on the initial push or stimulus that we impart. Based on this reasoning, consider a beam made out of iron and an otherwise identical beam made out of aluminum. How should the natural frequency of vibration of these two beams compare

Physics
1 answer:
Gekata [30.6K]3 years ago
3 0

Answer:

(b) In ideal condition we neglect mass of spring but in real springs mass of spring adds another factor to its time period.

since we are adding a factor of mass to the system, and frequency being inversely proportional to squared root of mass, we can come to a general conclusion that it effectively reduces the natural frequency .

Explanation:

kindly check the attachment for explanation.

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Please i need detailed explanation​
zubka84 [21]

Answer:

2Micro Farahds

Explanation:

Its in the picture.

I Hope it helps.

4 0
2 years ago
A container with volume 1.64 L is initially evacuated. Then it is filled with 0.226 g of N2N
vaieri [72.5K]

Answer:

0.015 atm

Explanation:

The pressure of the gas can be calculated using Ideal Gas Law:

p = \frac{nRT}{V}

<u>Where:</u>

n: is the number of moles of the gas

R: is the gas constant = 0.082 L*atm/(K*mol)

V: is the volume of the container = 1.64 L

T: is the temperature

We need to find the number of moles and the temperature. The number of moles is:

n = \frac{m}{M}

<u>Where:</u>

M: is the molar mass of the N₂ = 14.007 g/mol*2 = 28.014 g/mol

m: is the mass of the gas = 0.226 g

n = \frac{0.226 g}{28.014 g/mol} = 8.07 \cdot 10^{-3} moles

Now, the temperature can be found using the following equation:

v_{rms} = \sqrt{\frac{3RT}{M}}    

<u>Where:</u>

R: is the gas constant = 0.082 L*atm/K*mol = 8.314 J/K*mol

v_{rms}: is the root-mean-square speed of the gas = 182 m/s

By solving the above equation for T, we have:

T = \frac{v_{rms}^{2}*M}{3R} = \frac{(182 m/s)^{2}*28.014 \cdot 10^{-3} Kg/mol}{3*8.314 J K^{-1}mol^{-1}} = 37.20 K        

Finally, we can find the pressure of the gas:

p = \frac{nRT}{V} = \frac{8.07 \cdot 10^{-3} mol*0.082 L*atm* K^{-1}*mol^{-1}*37.20 K}{1.64 L} = 0.015 atm

Therefore, the pressure of the gas is 0.015 atm.

I hope it helps you!

8 0
3 years ago
Why copper is classified as a metal.
AveGali [126]

Answer:

It is classified as a metal because of its high ductility, malleability, thermal and electrical conductivity and resistance to corrosion. Copper is a mineral used for our everyday use. It is limited and we have to use it wisely because it is not evenly distributed everywhere and also everyone needs it. Copper is element number 29 on the Periodic Table of Elements. It is also used it electronics and wiring.

hope this helped

:)

8 0
3 years ago
you stand on a straight desert road at night and observe a vehicle approaching. this vehicle is equipped with two small headligh
inysia [295]

Answer:

The distance that you marginally able to discern that there are two headlights rather than a single light source is 6.084 km

Explanation:

Given:

d = distance = 0.679 m

λ = wavelength of the light = 537 nm = 537x10⁻⁹m

dp = pupil diameter = 4.81 mm = 0.00481 m

Question: What distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source, dx = ?

For the separation of the peak from the central maximum it is:

sin\theta =\frac{\lambda }{d_{p} } =\frac{537x10^{-9} }{0.00481} =1.116x10^{-4}

In this case, the two small sources of the headlights have the same angle as the images that form inside the eye

d_{x} =\frac{d}{sin\theta } =\frac{0.679}{1.116x10^{-4} } =6.084x10^{3} m=6.084km

7 0
3 years ago
There are two identical small metal spheres with charges 38.9 µC and −27.6399 µC. Thedistance between them is 6 cm. The spheres
Greeley [361]

Answer:

2683.3N

Explanation:

According to coulombs law which states that "the force of attraction existing between two charge q1 and q2 is directly proportional to the product of the charges and inversely proportional to the square of the distance (d) between them. Mathematically |F|= k|q1| |q2| /d² where;

F is the force of attraction between the charges

q1 and q2 are the charges

d is the distance between them

k is the coulombs constant

Given |q1|= 38.9 × 10^-6C and |q2| = 27.6399×10^-6C d = 6cm = 0.06m

k = 8.98755 × 109 Nm² /C²

Substituting the given data's in the equation we have;

|F| = 8.98755 × 10^9×38.9×10^-6×27.6399×10^-6/0.06²

|F| = 9.66/0.06²

|F| = 9.66/0.0036

|F| = 2683.3N

The magnitude of the force will be 2683.3N

Note that the modulus of the charges changes negative value of q2 to positive value. The opposite signs of the charges doesn't affect the final calculation, it only tells the force of attraction or repulsion between the charges. Since they are unlike charges, they will attract each other in the field.

4 0
3 years ago
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