All categories

3 years ago

A string is stretched to a length of 376 cm and

Physics

1 answer:

sergey [27]3 years ago

7 0

Answer: 53.09Hz

Explanation:

The fundamental frequency of an ideal taut string is:

Fn= n/2L(√T/μ)

Where:

F= frequency per second (Hz)

T= Tension of the string (cm/s sqr)

L= Length of the string (cm)

μ= Linear density or mass per unit length of the string in cm/gm

√T/μ= square root of T divided by μ

It is important to note:

Note: Typically, tension would be in newtons, length in meters and linear density in kg/m, but those units are inconvenient for calculations with strings. Here, the smaller units are used.

F1= 1/2(376cm)(0.01/1) × (√574/(0.036g/cm)(0.1kg/m÷1g/cm)

F1= 0.1329 × 399.30

= 53.09Hz

You might be interested in

Did the bird population appear to benefit from the elimination of the rattlesnakes

Maurinko [17]

Yes because there's less rattlesnakes to eat them

~~~hope this helps~~~
~~have a beautiful day~~
~davatar~

7 0

4 years ago

Tarzan, whose mass is 75 kg, is running from a cheetah. Tarzan is moving at 5 m/s when he grabs onto a hanging vine. How high of

Fiesta28 [93]

Answer:

Explanation:

His Kinetic energy = 1/2 m v^2

v = 5 m/s

m = 75 kg

Ke = 1/2 75 * 5^2

Ke = 937.5 Joules

This will be converted to PE when he reaches the maximum height he reaches. In other words KE = PE

PE = m * g * h

m = 75

g = 9.81

h = ?

PE = 937.5

937.5 = 75 * 9.81 * h

937.5 = 735.75 * h

937.5/735.75 = h

h= 1.27 meters

6 0

3 years ago

Why are people so rude???

DanielleElmas [232]

Answer:

I.D.K but the same measure they use to judge will be used to judge them

Explanation:

6 0

3 years ago

Read 2 more answers

A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate

yaroslaw [1]

Answer:

The maximum error is $\Delta \eta = 2032.9$

Explanation:

From the question we are told that

The length is $l = 1\ m$

The radius is $r = 0.002 \pm 0.0002 \ m$

The pressure is $P = 4 *10^{5} \ \pm 1750$

The rate is $v = 0.5*10^{-9} \ m^3 /t$

The viscosity is $\eta = \frac{\pi}{8} * \frac{P * r^4}{v}$

The error in the viscosity is mathematically represented as

$\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v$

Where $\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}$

and $\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}$

and $\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}$

$\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v]$

substituting values

$\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ]$