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ale4655 [162]
2 years ago
14

A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What wil

l be the wavelength if the tension is decreased to 600 N and the frequency is kept constant
Physics
1 answer:
mash [69]2 years ago
5 0

Answer:

The wavelength will be 33.9 cm

Explanation:

Given;

frequency of the wave, F = 1200 Hz

Tension on the wire, T = 800 N

wavelength, λ = 39.1 cm

F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}

Where;

F is the frequency of the wave

T is tension on the string

μ is mass per unit length of the string

λ is wavelength

\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu =  \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\

when the tension is decreased to 600 N, that is T₂ = 600 N

T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2  = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2  = \sqrt{0.11466} \\\\\lambda _2  =0.339 \ m\\\\\lambda _2  =33.9  \ cm

Therefore, the wavelength will be 33.9 cm

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4 0
2 years ago
PLEASEEE HELPPP!!!
Alja [10]

Given :

A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m.

The force of friction between the refrigerator and the floor is 230 N.

To Find :

How much work has been performed by the mover on the refrigerator.

Solution :

Since, refrigerator is moving with constant velocity.

So, force applied by the mover is also 230 N ( equal to force of friction ).

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3 0
3 years ago
A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area
weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A   (N/m²)

P1=P2

\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }

F_{2} = \frac{F_{1}*A_{2}  }{A_{1}}  Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight  (W₂) can the large piston support

We replace data in the equation (1)

F_{2} = \frac{(500)*(40) }{2}

F₂ = 10000 N

W₂= F₂= 10000 N

6 0
3 years ago
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