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3 years ago

Real springs have mass. How will the true period andfrequency

Physics

1 answer:

Ad libitum [116K]3 years ago

4 0

Explanation:

An perfect mass less spring, attached at one end and with a free mass attached at the other end, will have a distinct frequency of oscillation depending on its constant spring and mass. On the other hand, a spring with mass along its length will not have a characteristic frequency of oscillation.

Alternatively, based on its spring constant and mass per length, it will now have a wave Speed. It would be possible to use all wavelengths and frequencies, as long as the component fλ= S, where S is the spring wave size. If that sounds like longitudinal waves, like solid sound waves.

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Sub this channel pls... frreee points too.. ☺

Sunny_sXe [5.5K]

Answer:

I have sub to your channel :)

6 0

2 years ago

Read 2 more answers

a vertical solid steel post 29cm in diameter and 2.0m long is required to support a load of 8200kg, ignore the weight of the pos

erik [133]

Answer:

The stress is $\sigma = 1.218*10^{6} \ N/m^2$

Explanation:

From the question we are told that

The diameter of the post is $d = 29 \ cm = 0.29 \ m$

The length is $L = 2.0 \ m$

The weight of the loading mass

Generally the radius of the post is mathematically represented as

$r = \frac{0.29}{2}$

=> $r = 0.145 \ m$

Generally the area of the post is

$A = \pi r^2$

=> $A = 3.14 * 0.145 ^2$

=> $A = 0.066 \ m^2$

Generally the weight exerted by the load is mathematically represented as

$F = m * g$

=> $F = 8200 * 9.8$

=> $F = 80360 \ N$

Generally the stress is mathematically represented as

$\sigma = \frac{F}{A}$

=> $\sigma = \frac{80360 }{0.066}$

=> $\sigma = 1.218*10^{6} \ N/m^2$

7 0

2 years ago

Clouds in our atmosphere cause us to see the different phases of the moon.

krok68 [10]

Answer:

False

Explanation:

8 0

3 years ago

Read 2 more answers

What is the only possible value of ml for an electron in an s orbital?

Archy [21]

Answer:

zero

Explanation:

$m_l$ is the magnetic quantum number.

The only possible value for the magnetic quantum number for an electron in an s orbital is 0.

The first three quantun numbers are:

n: principal quantum number. It may have positive integer values: 1, 2, 3, 4,5, 6, 7, ...

$l$ : Azimuthal or angular momentum quantum number. It may have integer values from 0 to n - 1.

This quantum number is related to the type (or shape) of the orbital:

For s orbitals $l=0$

For p orbitals $l=1$

For d orbitals $l=2$

For f orbitals $l=3$

In this case, it is an s orbital, so we have $l=0$ .

$m_l$ , the third quantum number can have integer values ${from-l}$ to ${+l}$

Since, for the s orbitals $l=0$ , the only possible value for ${m_l}$ is zero.

4 0

3 years ago

A 1300 kg car traveling with a speed of 3.5 m/s executes a turn with a 8.5 m radius of curvature.

Y_Kistochka [10]

Answer:

1.4 m/s/s (2.s.f)

Explanation:

The formula for centripetal acceleration is:

$a=\frac{v^{2} }{r}$ , where v is velocity and r is the radius.

In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:

$a=\frac{3.5^{2} }{8.5} \\a=1.4$

Therefore the centripetal acceleration of the car is 1.4m/s/s. (2.s.f)

Hope this helped!

7 0

2 years ago