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Varvara68 [4.7K]
1 year ago
15

An Ethernet cable is 4.00m long. The cable has a mass of 0.200kg . A transverse pulse is produced by plucking one end of the tau

t cable. The pulse makes four trips down and back along the cable in 0.800 s. What is the tension in the cable?
Physics
1 answer:
Mashcka [7]1 year ago
4 0

Based on the length of the Ethernet cable and the mass, the tension in the cable can be found to be 80 N.

<h3>How much tension is in the cable?</h3>

The tension in the cable can be found as:

= 4 x mass x length x frequency

Solving for the frequency is:

= 1 / (0.800 / 4)

= 1 / 0.20

= 5.0 Hz

The tension is therefore:

= 4 x 0.20 x 4.00 x 5

= 80N

Find out more on tension at brainly.com/question/14336853

#SPJ4

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Complete Question:

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experiences the smallest drop in temperature, and which one experiences the largest drop? Sample A: 4.0 kg of water [c = 4186 J/(kg·C°)] Sample B: 2.0 kg of oil [c = 2700 J/(kg·C°)] Sample C: 9.0 kg of dirt [c = 1050 J/(kg·C°)]

Answer:

A. Smallest B. Largest.

Explanation:

Assuming no heat exchange except for the heat removed from any sample (which we know is the same for the three ones), and that the process is done using only conduction, we can use the equation that relates the heat lost or gained by one object, with the mass of the object and the consequent change in temperature, as follows:

Q = c*m*ΔT, where c, is a proportionality constant called specific heat, which is different for each material.

As we know that the heat removed is the same for the three samples, we can equate the right sides of the equation for each sample, as follows:

cw*mw*ΔTw = co*mo*ΔTo = cd*md*ΔTd

Replacing by the givens, we have:

4.0 kg. 4,186 J/kgºC*ΔT(ºC) = 2.0 kg*2,700 J/kgºC*ΔT(ºC) =9.0kg*1,050J/kgºC*ΔT(ºC)

As the three expressions must be equal each other, it's clear that the unknown term (the drop in temperature) must compensate the product of the mass times the specific heat.

This product is the following for the three samples:

Water: 4.0 kg*4,186 J/kgºC = 16,744 J/ºC

Oil : 2.0 kg*2,700 J/kgºC    = 5,400 J/ºC

Dirt: 9.0 * 1,050 J/kgºC        = 9,450 J/ºC

Clearly, we see that in order to keep the heat exchange equations equal each other, the water must suffer the smallest drop in temperature, and the oil must experience the largest one.

So, the sample A experiencies the smallest drop in temperature, and sample B does the largest one.

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