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Lorico [155]
3 years ago
7

A 13 kg hanging sculpture is suspended by a 95-cm-long, 5.0 g steel wire. When the wind blows hard, the wire hums at its fundame

ntal frequency. What is the frequency of the hum

Physics
2 answers:
Valentin [98]3 years ago
8 0

Explanation:

Below is an attachment containing the solution.

Artyom0805 [142]3 years ago
3 0

Answer:

f=81.96 \ Hz

Explanation:

Givens

L=95cm

m_{sculpture} =13kg

m_{wire}=5g

The frequency is defined by

f=\frac{v}{\lambda}

Where v is the speed of the wave in the string and \lambda is its wave length.

The wave length is defined as \lambda = 2L = 2(0.95m)=1.9m

Now, to find the speed, we need the tension of the wire and its linear mass density

v=\sqrt{\frac{T}{\mu} }

Where \mu=\frac{0.005kg}{0.95m}= 5.26 \times 10^{-3} and the tension is defined as T=m_{sculpture} g=13kg(9.81 m/s^{2} )=127.53N

Replacing this value, the speed is

v=\sqrt{\frac{127.53N}{5.26 \times 10^{-3} } }=155.71 m/s

Then, we replace the speed and the wave length in the first equation

f=\frac{v}{\lambda}\\f=\frac{155.71 m/s}{1.9m}\\ f=81.96Hz

Therefore, the frequency is f=81.96 \ Hz

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Here are three pairs of initial and final positions, respectively, along an x-axis. Which pair gives a
Sladkaya [172]

Answer:

(b)-3 m,-7 m

Explanation:

Displacement is a vector quantity that connects the initial position of an object to its final position after a certain motion.

Mathematically, displacement can be written as:

d=x_f - x_i

where

x_f is the final position

x_i is the initial position

The direction of this vector is from the initial position to the final position.

In order to calculate the displacement of an object, therefore, it is necessary to correctly take into account the sign of each motion, which gives the direction of each motion.

Let's analyze the three cases:

(a) -3 m, +5 m;

In this case, the object moves 3 m in the negative direction first, and then 5 meters in the positive direction; so the net displacement is

d = -3 + 5 = +2 m

(b)-3 m,-7 m;

In this case, the object moves 3 m in the negative direction first, and then 3 meters in the negative direction; so the net displacement is

d = -3 - 7 = -10 m

(c) 7 m, -3 m?

In this case, the object moves 7 m in the positive direction first, and then 3 meters in the negative direction; so the net displacement is

d = +7 + (-3) = +4 m

So, the only situation in which we have a net negative displacement is situation b).

6 0
3 years ago
Which of the following four circuit diagrams best represents the experiment described in this problem?
Valentin [98]

We don't see any circuit diagrams.  

This worries us for a few seconds, until we realize that we don't know anything about the experiment described in the problem either, so we don't have to worry about it at all.

6 0
3 years ago
A fire hose has an inside diameter of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of 1.
inessss [21]

Answer:

The Reynolds numbers for flow in the fire hose.

Explanation:

Given that,

Diameter = 6.40 cm

Rate of flow = 40.0 L/s

Pressure P=1.62\times10^{6}\ N/m^2

We need to calculate the Reynolds numbers for flow in the fire hose

Using formula of rate of flow

Q=Av

v=\dfrac{Q}{A}

Where, Q = rate of flow

A = area of cross section

Put the value into the formula

v=\dfrac{40.0\times10^{-3}}{3.14\times(3.2\times10^{-2})^2}

v=12.44\ m/s

We need to calculate the Reynolds number

Using formula of the Reynolds number

n_{R}=\dfrac{2\rho\times v\times r}{\eta}

Where, \eta =viscosity of fluid

\rho =density of fluid

Put the value into the formula

n_{R}=\dfrac{2\times100\times12.44\times3.2\times10^{-2}}{1.002\times10^{-3}}

n_{R}=7.945\times10^{5}

Hence, The Reynolds numbers for flow in the fire hose.

3 0
3 years ago
Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis wit
fgiga [73]

Answer:

The effect of this rotation on the person on the surface is the sky is moving, like an apparent diurnal motion.

Explanation:

<em>For an observer at a fixed position on Earth, the rotation of earth makes it appear as if the sky is revolving around the earth. In other words, if a person is standing for long enough in a field at night, it looks like the sky is moving, not the person. This motion is called "apparent diurnal motion." </em>

<em>Diurnal means having to do with a day, in a sense of a 24- hour period.</em>

5 0
3 years ago
A fan blade is rotating with a constant angular acceleration of 11.5 rad/s2. At what point on the blade, as measured from the ax
vichka [17]

Answer:

Alpha = ω^2 R    where R is radius of blade

g = w^2 r      where r is distance from center

ω^2 R = 11.5 ω^2 r

R / r = 11.5 / 9.8 = 1.17

Or r = .852 R

Since the angular acceleration depends on both R and ω it seems that one can only get r as it depends on R

7 0
2 years ago
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