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Dmitry_Shevchenko [17]
1 year ago
15

A ski resort consists of a few chairlifts and several interconnected downhill runs on the side of a mountain, with a lodge at th

e bottom. The chairlifts are analogous to batteries, and the runs are analogous to resistors. Describe how two runs can be in series. Describe how three runs can be in parallel. Sketch a junction between one chairlift and two runs. State Kirchhoff's junction rule for ski resorts. One of the skiers happens to be carrying a skydiver's altimeter. She never takes the same set of chairlifts and runs twice, but keeps passing you at the fixed location where you are working. State Kirchhoff's loop rule for ski resorts.
Physics
1 answer:
Aleks04 [339]1 year ago
7 0
  1. A sketch of the junction between one chairlift and two runs is shown in the image attached below.
  2. For this ski resort, it ultimately implies that the total number of lifts that are used in any closed loop would be equal to zero in accordance with Kirchhoff's loop rule.

<h3>What is Kirchhoff's loop rule?</h3>

Kirchhoff's loop rule is also referred to as Kirchhoff's second law or Kirchhoff's voltage law, and it states that the algebraic sum of all the electric potential differences around any closed loop is equal to zero.

<h3>How to connect two runs in series?</h3>

Since the chairlifts are analogous to batteries, and the runs are analogous to resistors, the two runs would represent resistors that are connected in series with the lift, which represents the battery. Similarly, these three runs can also be connected in parallel by altering the positions of the runs with the battery.

For this ski resort, it ultimately implies that the total number of lifts that are used in any closed loop would be equal to zero in accordance with Kirchhoff's loop rule.

Read more on Kirchhoff's loop rule here: brainly.com/question/15003023

#SPJ4

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Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.
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a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

a)

We can solve this first part of the problem by applying the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

f=\frac{R}{2}

For this mirror, R = 20 cm, so its focal length is

f=\frac{20}{2}=+10 cm (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm

b)

We can solve this part by using the magnification equation:

M=-\frac{y'}{y}=\frac{q}{p}

where

y' is the size of the image

y is the size of the object

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p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm

and the negative sign means that the image is inverted.

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