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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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Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other.
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Incomplete question.The complete question is attached below as screenshot along with figure

Answer:

F=6.00*10^{-6}N

Force is repulsive

Explanation:

Given data

Current I₁=5.00A

Current I₂=2.00A

Length L=1.20 m

Radius r=0.400m

To find

Force F

Solution

As the force is repulsive because currents are in opposite direction

From repulsive force we know that:

F=\frac{u_{o}I_{1}I_{2}L}{2\pi r}

Substitute the given values

F=\frac{u_{o}(5.00A)(2.00A)(1.20m)}{2\pi (0.400m)}\\ F=6.00*10^{-6}N

4 0
3 years ago
A heavy object and a light object are dropped from the same height. If we neglect air resistance, which will hit the ground firs
Maksim231197 [3]

Answer:

None, both objects will hit ground at the same time.

Explanation:

  • Assuming no air resistance present, and that both objects start from rest, we can apply the following kinematic equation for the vertical displacement:

        \Delta h = \frac{1}{2}*g*t^{2}  (1)

  • As the left side in (1) is the same for both objects, the right side will be the same also.
  • Since g is constant close to the surface of the Earth, it's also the same for both objects.
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Q=8.0 nC=8.0 \cdot 10^{-9} C is the charge of the bead
r=3.0 cm=0.03 m is the distance at which we calculate the field.

Using these data, we find:
E=(8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{(8.0 \cdot 10^{-9} C}{0.03 m)^2}=8.0 \cdot 10^4 N/C
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When the grasshoppers vertical velocity is exactly zero.
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6 0
3 years ago
Read 2 more answers
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