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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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Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay
Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

mass, m = 0.1 kg

Amplitude, A = 5 cm = 0.05 m

Let the angular frequency is w.

w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s

The maximum velocity is

v_{max} = w A\\\\v_{max} = 31.6\times 0.05 = 1.58 m/s

8 0
3 years ago
If yall friend me ill give lots of points like this in the future
Ne4ueva [31]

Answer:

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3 0
2 years ago
The momentum of an object is determined by the object’s mass and velocity. Which object has more momentum?
Gnesinka [82]

Answer:

You need to give the options but the formula is p=mv

Explanation:

7 0
2 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
A runner traveled 15 kilometers north then backtracked 11 kilometers south before stopping. His resultant displacement was
Serhud [2]

<u>Answer:</u>

 Resultant displacement is 4 km north.

<u>Explanation:</u>

    Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.

   A runner traveled 15 kilometers north then backtracked 11 kilometers south before stopping,

   So initial displacement, 15 kilometers north = 15 j km

        Second displacement, 11 kilometers south = -11 j km

   Total displacement = 15 j -11 j = 4 j km

   Total displacement is 4 km north.

4 0
3 years ago
Read 2 more answers
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