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S_A_V [24]
3 years ago
14

A 240 cm length of string has a mass of 2.5 g. It is stretched with a tension of 8.0 N between fixed supports. (a) What is the w

ave speed for this string? (b) What is the lowest resonant frequency of this string?
Physics
1 answer:
kotykmax [81]3 years ago
3 0

Answer:

a.87.6 m/s

b.18.25 Hz

Explanation:

the equation for the wave speed is expressed as

v=\sqrt{\frac{Tl}{m}}\\

where v is the speed,

           T is the tension in Newton

           l is the length

and      m is the mass

Now since

T=8.0N\\m=2.5g=0.0025kg\\l= 240cm=2.40cm\\

by substituting values into the equation, we have

v=\sqrt{\frac{8*2.40}{0.0025} } \\v=87.6m/s\\

b. the expression for the frequency is giving as

Frequency,f=\frac{v}{2l} \\f=\frac{87.6}{2*2.40} \\f=18.25Hz\\.

Note we 2L as the wavelength because we solving for the fundamental frequency as stated in the question.

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Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 2.88 kg and rotate with
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Answer:

(a) K_{small}=4839.3J

(b) K_{larger}=17406.4J

Explanation:

Given data

The angular velocity of two cylinders ω=257 rad/s

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The radius of small cylinder r₁=0.319 m

The radius of larger cylinder r₂=0.605 m

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The rotational kinetic energy of the cylinder is given by:

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So

K_{small}=\frac{1}{4}(2.88kg)(0.319)^2(257rad/s)^2 \\K_{small}=4839.3J

For Part (b)

K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr_{2}^2)w^2

Substitute the given values

K_{larger}=\frac{1}{4}mr_{2}^2w^2\\ K_{larger}=\frac{1}{4}(2.88kg)(0.605m)^2(257rad/s)^2\\ K_{larger}=17406.4J

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3 years ago
A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o
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Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

P=\rho \times g \times h

Here

  • P is the pressure which is given as 68 kPa.
  • ρ is the density of the oil whose SG is 0.92. It is calculated as

                                       \rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\

  • g is the gravitational constant whose value is 9.8 m/s^2
  • h is the height which is to be calculated

                                        P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m

So the height of column is 7.54m

a-3

By the relation of volume and density

M=\rho \times V

Here

  • ρ is the density of the oil which is 920 kg/m^3
  • V is the volume of cylinder with diameter 16m calculated as follows

                             V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3

Mass is given as

                             M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg

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Pressure variation is given as

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Now corrected pressure is as

P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa

Finding the value of height for this corrected pressure as

P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m

The original height of column is 5.98m

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