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3 years ago

Which best explains how fiber-optic technology has improved communication?

Physics

2 answers:

Gelneren [198K]3 years ago

8 0

Answer:

B. It has allowed for faster transmission of Internet signals.

Explanation:

i took the test on engenuity

Kobotan [32]3 years ago

7 0

Answer:

Its B

Explanation:

: )

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The Atwood’s machine shown consists of two blocks of mass m1 and m2 that are connected by a light string that passes over a pull

Talja [164]

(A) For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

(D) For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.

<h3 /><h3>The given parameters:</h3>

Mass of block 1 = m1
Mass of block 2, = m2
Height of block 1 above the ground, = h1
Height of block 2 above the ground = h2

The total initial mechanical energy of the two block system is calculated as follows;

$m_1gh_1 + \frac{1}{2} m_1v_1_i^2 = m_2gh_2 + \frac{1}{2} m_2v_2_i^2\\\\m_1gh_1 + 0 = m_2gh_2 + 0\\\\m_1gh_1 = m_2gh_2\\\\m_1gh_1 - m_2gh_2 = 0$

When the block m2 reaches the ground the block m1 attains maximum height and the total mechanical energy at this point is given as;

$m_1g(h_1 + h_2) + K.E_1 = \frac{1}{2}m_2v_{max}^2 + P.E_2\\\\m_1g(h_1 + h_2 ) -PE_2 = \frac{1}{2}m_2v_{max}^2 - K.E_1\\\\m_1g(h_1 + h_2 ) - 0= \frac{1}{2}m_2v_{max}^2 - 0\\\\m_1g(h_1 + h_2 ) = \frac{1}{2}m_2v_{max}^2\\\\W = \frac{1}{2}m_2v_{max}^2$

Thus, we can conclude the following before the block m2 reaches the ground;

For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

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3 years ago

Please elp! will mark BRAINLIEST

artcher [175]

Answer:

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Explanation:

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4 years ago

How lift force works in CER format

Step2247 [10]

The really good answer is 30cm CER format

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3 years ago

If you calculate W, the amount of work it took to assemble this charge configuration if the point charges were initially infinit

sp2606 [1]

Answer:

W = 0×(kq2L)

Explanation:

We know that the work to assemble a charge configuration of two charges a distance r from each other is simply W = kq2/r

If we want to assemble three charges A, B, and C. It's necessary to consider the distances between them

WABC = kq2/(rAB + rAC + rBC)

So, to assemble four charges A, B, C, & D....

WABCD = kq2/(rAB + rAC + rAD + rBC + rBD + rCD)

Considering a square charge configuration with sides L, such as in figure attached A, B, & C are positive & D is negative

rAB = L

rAC = L√2

rAD = L (-)

rBC = L

rBD = L√2 (-)

rCD = L (-)

⇒ W = kq2/(L + L√2 + (-L) + L + (-L√2) + (-L)

⇒ ∴ W = 0 × (kq2/L)

This way, working through each option...

(a)

The positive charges are equidistant from each other at a distance of L.

rAB = L

rAC = L

rAD = ½L⋅sin(60) (-)

rBC = L

rBD = ½L⋅sin(60) (-)

rCD = ½L⋅sin(60) (-)

Wa = kq2/(3L - (3/2)L⋅(0.866))

⇒ ∴ Wa = (1/1.7) × (kq2/L) = (0.5879)× (kq2/L)

(b)

rAB = L

rAC = 2L

rAD = 3L (-)

rBC = L

rBD = 2L (-)

rCD = L (-)

Wb = kq2/(4L - 6L)

⇒ ∴ Wb = (-1/2) × (kq2/L) = (-0.5)× (kq2/L)

(c)

The factor doesn't matter, so Wc = 0 × (kq2/L)

In this case, the greater work is actually the less work. Therefore, the positive work represents the amount of work the system actually exhibits, that we don't have to do. If there is negative work, we have to make up that work in order to place the charges as desired.

This way, charge configuration (a) requires the least amount of work.

5 0

3 years ago

A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of co

Genrish500 [490]

Answer: When 1.0kg of aluminium block is used, the final temperature of the mixture will be T = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be= 147.1∘C

Explanation:

H = mcΘ

heat lost by block = heat gained by water

m₁c₁Θ₁ = m₂c₂Θ₂ where m₁ is mass of aluminium, m₂ is mass of water, c₁ is cAluminium, c₂ is cWater, Θ₁ is temperature change for aluminium, Θ₂ is temperature change for water.

0.5*900*(200-20) = m₁*4186*(20-0)

m₁ = 450*180/83270

<em>when 1.0kg of aluminium block is used, the final temperature of the mixture will be </em><em>T</em>

heat lost by block = heat gained by water

1.0*900*(200-T) = 0.973*4186*(T-0)

180000 - 900T = 4073T

4973T = 180000

T = 180000/4973 = 36.2∘C

<em>If 1.0kg copper block is used, T of the mixture will be</em>

heat lost by block = heat gained by water

1.0*387*(200-T) = 0.973*4186*(T-0)

77400 - 387T = 4073T

4460T = 77400

T = 77400/4460 = 17.4∘C

<em>heat lost by block = heat gained by water + heat used in melting ice to form water at 0∘C</em>

heat used in melting 0.1kg of ice, H = ml, where l= 33600J/Kg

0.5*900*(200-T) = 0.1*4186*(T-0) + 0.1*33600J/Kg

90000 - 450T = 418.6T + 33600

418.6T + 450T = 90000 - 33600

868.6T = 56400

T = 56400/868.6 = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be

0.5*900*(200-T) = 0.025*4186*(T-0) + 0.025*33600J/Kg

90000 - 450T = 104.65T + 8400

104.65T + 450T = 90000 - 8400

554.65T = 81600

T = 81600/554.65 = 147.1∘C

7 0

3 years ago