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3 years ago

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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

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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An old-fashioned LP record rotates at 33 1/3 RPM

Ksenya-84 [330]

Answer:

Part a) $\frac{5}{9}\ \frac{rev}{sec}$

Part b) $\frac{9}{5}\ \frac{sec}{rev}$

Explanation:

Part a) what is its frequency, in rev/s

we have that

An old-fashioned LP record rotates at 33 1/3 RPM

so

$33\frac{1}{3}\ \frac{rev}{min}$

Convert mixed number to an improper fraction

$33\frac{1}{3}\ \frac{rev}{min}=\frac{33*3+1}{3}=\frac{100}{3}\ \frac{rev}{min}$

Remember that

$1\ min=60\ sec$

Convert rev/min to rev/sec

$\frac{100}{3}\ \frac{rev}{min}=\frac{100}{3}(\frac{1}{60})=\frac{100}{180}\ \frac{rev}{sec}$

Simplify

$\frac{5}{9}\ \frac{rev}{sec}$

Part b) what is it period, in seconds

we know that

The period is the reciprocal of the frequency

therefore

the frequency is

$\frac{9}{5}\ \frac{sec}{rev}$

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3 years ago

An indicator of average kinetic energy is _____. pressure volume diffusion temperature

slega [8]

An indicator of average kinetic energy is temperature. Temperature is directly proportional to Kinetic energy of the molecules of an element.

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1. Bone has a Young’s modulus of about

Blababa [14]

#1

As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

$strain = 9.33 \times 10^{-3}$

now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

$\Delta L = 5.04 \times 10^{-3} m$

$\Delta L = 5.04 mm$

#2

As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

#3

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

$B = 2.3 \times 10^9 N/m^2$

now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

$\rho = 1185.3 kg/m^3$

#4

as we know that pressure changes with depth as per following equation

$P = P_o + \rho g h$

here we know that

$P = 3 P_0$

now we will have

$3P_0 = P_0 + \rho g h$

$2P_0 = \rho g h$

$2(1.01 \times 10^5) = 1025 (9.81)(h)$

here we will have

$h = 20.1 m$

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

$mg = \rho_1V_1g + \rho_2V_2g$

$A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)$

$46.6 = 43.89 - 922x + 1000x$

$x = 3.48 cm$

so it is 3.48 cm below the interface

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Does anyone know this???

Gelneren [198K]

Answer:

C

Explanation:

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When an experiment is replicated, how should the results of the two experiments compare?

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When the experiment is replicated, this means the conduction of the second experiment, should be related, or similar, to your first results.

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