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3 years ago

What is the name of the specification that indicates the frequencies that are dedicated for data transmission over cable lines?

1 answer:

mihalych1998 [28]3 years ago

4 0

The answer to this question is the term
DOCSIS. A DOCSIS or Data Over Cable Service Interface Specification is a telecommunication standard or interface where in an internet was being provided by the use of cables. The advantage of using DOCSIS is that the speed of the internet is faster using this kind of interface.

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Why might acting happy actually make one happy

dexar [7]

Because they are mentally trying to extinguish the negative things in their lives and focus on the positive things

5 0

3 years ago

How was newton's laws used to solve problems on apollo 13 ship<br> Explain in two full paragraphs

MaRussiya [10]

Answer:

actually ships are made in newtons third law of motion.it states to every action there is equal and opposite reaction. curved is made in downwards to maintain upthrust and to made balance.

actually it prevents ships from drowning and to move with a heavy mass.

3 0

2 years ago

1. A 14-cm tall object is placed 26 cm from a converging lens that has a focal length of 13 cm.

AURORKA [14]

Answer:

a) Please find attached the required drawing of light passing through the lens

By the use of similar triangles;

The image distance from the lens = 26 cm

The height of the image = 14 cm

c) The image distance from the lens = 26 cm

The height of the image = 14 cm

Explanation:

Question;

a) Determine the image distance and the height of the image

b) Calculate the image position and height

The given parameters are;

The height of the object, h = 14 cm

The distance of the object from the mirror, u = 26 cm

The focal length of the mirror, f = 13 cm

The location of the object = 2 × The focal length

Therefore, given that the center of curvature ≈ 2 × The focal length, we have;

The location of the object ≈ The center of curvature of the lens

The diagram of the object, lens and image created with MS Visio is attached

From the diagram, it can be observed, using similar triangles, that the image distance from the lens = The object distance from the lens = 26 m

The height of the image = The height of the object - 14 cm

b) The lens equation is used for finding the image distance from the lens as follows;

$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$

Where;

v = The image distance from the lens

We get;

$v = \dfrac{u \times f}{u - f}$

Therefore;

$v = \dfrac{26 \times 13}{26 - 13} = \dfrac{26 \times 13}{13} = 26$

The distance of the image from the lens, v = 26 cm

The magnification, M =v/u

∴ M = 26/26 = 1, therefore, the object and the image are the same size

Therefore;

The height of the image = The height of the object = 14 cm.

5 0

2 years ago

Keesha conducts an experiment by pouring equal amounts of boiling water into four containers. The containers are the same size a

dybincka [34]

Option Z is correct. The stainless steel container was likely the hottest. Stainless steel is an excellent heat conductor because it quickly warms the substance.

<h3 /><h3>What are the qualities of stainless steel?</h3>

Stainless steel is an excellent heat conductor because it quickly warms the substance or allows heat to travel through it. Stainless steel is also corrosion-resistant.

Foam, glass, and plastic, on the other hand, are all poor heat and electrical conductors. As a result, they do not allow heat to travel through.

As a result, we may deduce that, among the available possibilities, the stainless steel container was most likely the hottest.

Hence Option Z is correct. The stainless steel container was likely the hottest.

To learn more about stainless steel refer to the link;

brainly.com/question/94280

5 0

2 years ago

A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does t

Ede4ka [16]

Answer:

(A) The maximum height of the ball is 40.57 m

(B) Time spent by the ball on air is 5.76 s

Explanation:

Given;

initial velocity of the ball, u = 28.2 m/s

(A) The maximum height

At maximum height, the final velocity, v = 0

v² = u² -2gh

u² = 2gh

$h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m$

(B) Time spent by the ball on air

Time of flight = Time to reach maximum height + time to hit ground.

Time to reach maximum height = time to hit ground.

Time to reach maximum height is given by;

v = u - gt

u = gt

$t = \frac{u}{g}$

Time of flight, T = 2t

$T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s$

As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.

v² = u² - 2gh

12² = 28.2² - 2(9.8)h

12² - 28.2² = - 2(9.8)h

-651.24 = -19.6h

h = 651.24 / 19.6

h = 33.23 m

Thus, at 33.23 m the speed will be 12 m/s

6 0

3 years ago