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3 years ago

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Two objects are maintained at constant temperatures, one hot and one cold. Two identical bars can be attached end to end, as in

part a of the drawing, or one on top of the other, as in part b. When either of these arrangements is placed between the hot and the cold objects for the same amount of time, heat Q flows from left to right. Find the ratio Qa/Qb.

1 answer:

irina1246 [14]3 years ago

7 0

Answer:

Qa/Qb = k/2×2k = 1/4

Explanation:

a situation is series and b situation is parallel.

let conductance of each plate = k

so net conductance in series = k/2

net conductance in parallel = 2k

so Qa/Qb = k/2×2k = 1/4

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In our solar system, the most likely planet (other than Earth) to have life on it is currently thought to be

VLD [36.1K]

Answer: Mars

Explanation:

Mars is a planet similar to the earth which the rest can't sustain life either because it's too hot or cold or too much gas maybe even toxic acid.

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3 years ago

Sheila weighs 60 kg and is riding a bike. Her momentum on the bike is 340 kg • m/s. The bike hits a rock, which stops it complet

Vikki [24]

Answer:

v₂ = 5.7 m/s

Explanation:

We will apply the law of conservation of momentum here:

$Total\ Initial\ Momentum = m_{1}v_{1} + m_{2}v_{2}\\$

where,

Total Initial Momentum = 340 kg.m/s

m₁ = mass of bike

v₁ = final speed of bike = 0 m/s

m₂ = mass of Sheila = 60 kg

v₂ = final speed of Sheila = ?

Therefore,

$340\ kg.m/s = m_{1}(0\ m/s) + (60\ kg)v_{2}\\v_{2} = \frac{340\ kg.m/s}{60\ kg}\\\\$

<u>v₂ = 5.7 m/s </u>

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3 years ago

What problem do refractor telescopes have that reflectors don't? Group of answer choices bad seeing light loss from secondary el

Paul [167]

chromatic aberration problem do refractor telescopes have that reflectors don't

<u>Explanation:</u>

Chromatic aberration is a phenom in which light rays crossing through a lens focus at various points, depending on their wavelength. Chromatic aberration is a dilemma in which lens or refracting, telescopes undergo from. The various image distances for the respective colors affect various image sizes for them.

This involves the creation of disturbing color fringes in the image. Chromatic aberration can be pretty well adjusted by the use of an achromatic doublet. Here, a positive biconvex lens is coupled with a negative lens placed backward with greater dispersion. Thus partly compensates for the chromatic aberration.

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3 years ago

The two measurements necessary for calculating average speed are

alisha [4.7K]

The correct answer is Option (C) distance and time

Explanation:

Average speed of any object is defined as the total distance that object travels over the time it takes to travel that distance. In other words, average speed is the total distance divided by the elapsed time.

$Average \thinspace Speed = \frac{Total \thinspace Distance}{Elapsed \thinspace Time}$

Therefore, as you can see in the above equation, the two measurements that are essential for the calculation of the average speed are the (total) distance and the (elapsed) time.

Hence, the correct option is C.

5 0

3 years ago

An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the po

viktelen [127]

Answer:

$1.232\times 10^{-5}\ T.$

Explanation:

<u>Given:</u>

Current through the wire, passing through the origin, $I_1 = 250\ A.$

Current through the wire, passing through the y axis, $r_y=1.8\ m.$ , $I_2 = 50\ A.$

According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to $\mu_o$ times the net current threading the loop.

$\oint \vec B \cdot d\vec l=\mu_o I.$

In case of a circular loop, the directions of magnetic field and the line element $d\vec l$ , both are along the tangent of the loop at that point, therefore, $\vec B\cdot d\vec l = B\ dl$ .

$\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\$

$\oint dl$ is the circumference of the Amperian loop = $2\pi r$

Therefore,

$B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.$

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_1 = 3.510\ m.$

The magnetic field at the given point due to this wire is given by:

$B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.$

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_2 = 5.310\ m.$

The magnetic field at the given point due to this wire is given by:

$B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.$

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at $r_r=-3.510\ m$ is given by

$B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.$

6 0

3 years ago