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Varvara68 [4.7K]
2 years 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]2 years 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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Suatu sistem gas berada didalam ruang yang fleksibel. Pada awalnya gas berada pada kondisi P1 = 1,5 × 10 pangkat 5 N/m2, T1 = 27
enot [183]

Answer:

16.00L

Explanation:

First you calculate the number of moles in the system:

PV=nRT\\\\n=\frac{PV}{RT}\\\\n=\frac{(1.5*10^5N/m^2)(12L)}{(0.082L.atm/mol.K)(300.15K)}=73134.16\ mol

To find the new volume of the system you use the following formula for an isobaric procedure:

T_2-T_1=\frac{P}{nR}(V_2-V_1)\\\\V_2=\frac{nR}{P}(T_2-T_1)+V_1\\\\V_2=\frac{(73134.16\ mol)(0.082L.atm/mol.K)}{1.5*10^5N/m^2}(400.15-300.15)K+12L\\\\V_2=16.00L

hence, the new volume is 16.00L

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4 years ago
A battery uses _________________ energy to generate _______________ energy.
Alisiya [41]

Uses chemical and generates electrical
7 0
4 years ago
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Does a ship float higher in the water of an inland lake or in the ocean? Why?
Ahat [919]

The weight of a ship is frequently called its "displacement" since that's the weight of the water that it uproots. It'll drift when it uproots a volume of water whose weight is break even with the weight of the ship  -- this can be the buoyant drive given by the water. New water in an inland lake features a littler density than that of ocean water. Hence, a larger volume of new water is vital to supply the same weight or buoyant force. This implies the ship will ride lower in an inland lake and will ride higher within the sea.

<h3>what is buoyant force?</h3>

When an object is set in a liquid, the liquid applies an upward force we call the buoyant force. The buoyant force comes from the weight applied to the question by the liquid. Since the weight increments as the profundity increments, the weight on the foot of an object is continuously bigger than the force on the best - consequently the net upward force. The buoyant force is present whether the question coasts or sinks.

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7 0
2 years ago
The given function represents the position of a particle traveling along a horizontal line. s(t) = 2t3 − 3t2 − 36t + 6 for t ≥ 0
igor_vitrenko [27]

1) The velocity of the particle is given by the derivative of the position. So, if we derive s(t), we get the velocity of the particle as a function of the time:

v(t)=s'(t)=(2t^3-3t^2-36t+6)'=6t^2-6t-36

2) The acceleration of the particle is given by the derivative of the velocity. So, if we derive v(t), we get the acceleration of the particle as a function of the time:

a(t)=v'(t)=(6t^2-6t-36)'=12t-6

8 0
4 years ago
The density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cubic
Drupady [299]

a) Density at 100 degrees: 1.34\cdot 10^4 kg/m^3

Explanation:

The density of mercury at 0 degrees is d=1.36\cdot 10^4 kg/m^3

Let's take 1 kg of mercury. Its volume at 0 degrees is

V=\frac{m}{d}=\frac{1 kg}{1.36\cdot 10^4 kg/m^3}=7.35\cdot 10^{-5} m^3

The formula to calculate the volumetric expansion of the mercury is:

\Delta V= \alpha V \Delta T

where

\alpha=180\cdot 10^{-6} K^{-1} is the cubic expansivity of mercury

V is the initial volume

\Delta T is the increase in temperature

In this part of the problem, \Delta T=100 C-0 C=100 C=100 K

So, the expansion is

\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(100 K)=1.3\cdot 10^{-6} m^3

So, the new density is

d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+1.3\cdot 10^{-6} m^3}=1.34\cdot 10^4 kg/m^3


b) Density at 22 degrees: 1.355\cdot 10^4 kg/m^3

We can apply the same formula we used before, the only difference here is that the increase in temperature is

\Delta T=22 C-0 C=22 C=22 K

And the volumetric expansion is

\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(22 K)=2.9\cdot 10^{-7} m^3

So, the new density is

d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+2.9\cdot 10^{-7} m^3}=1.355\cdot 10^4 kg/m^3


8 0
3 years ago
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