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marissa [1.9K]
3 years ago
5

A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N

, what is the fundamental frequency
Physics
1 answer:
Gre4nikov [31]3 years ago
5 0

Answer:

<h2>f₀ = 158.12 Hertz</h2>

Explanation:

The fundamental frequency of the string  f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.

V = \sqrt{\frac{T}{\mu} } where T is the tension in the string and  \mu is the density of the string

Given T = 600N and \mu = 0.015 g/cm  = 0.0015kg/m

V = \sqrt{\frac{600}{0.0015} }\\ \\V =  \sqrt{400,000}\\ \\V = 632.46m/s

The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.

L = 2m

Substituting the derived values into the formula f₀ = V/2L

f₀ = 632.46/2(2)

f₀ = 632.46/4

f₀ = 158.12 Hertz

Hence the fundamental frequency of the string is 158.12 Hertz

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When the temperature decreases the particals start to slow down.

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A 1250 kg car is pulling a 325 kb trailer. The car and trailer have an acceleration of 2,15m/sec2 together. Determine the net fo
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F= ma

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F= 698.75N

Under the reference system in which the direction of travel is the positive direction, the direction of the force will be positive.

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Answer:

w =  - 508.53 joules

q = - 3091.47 joules

Explanation:

Let us convert the time in hours into seconds

0.010* 3600\\= 36

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\delta E = p * \delta t

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

\delta E = - 100 * 36\\

\delta E = - 3600 Joules

Amount of work done by the system

w = - P * \delta V

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

w = - 1 * ( 5.92 -0.90)\\

w = -5.02 liter-atmospheres

Work done in Joules

- 5.02 * 101.3\\= 508.53Joules

q = \delta E - w\\

Substituting the given values we get -

q = - 3600 - (-508.53)\\q = - 3091.47

Thus

w =  - 508.53 joules

q = - 3091.47 joules

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Which of the following is most likely to be an observation made by a physiologist
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A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?
Pie

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

6 0
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