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Snowcat [4.5K]
3 years ago
14

Which of the following items can enter and exit an isolated system?

Physics
1 answer:
Mkey [24]3 years ago
7 0

Neither energy nor matter can enter and exit an isolated system.

Explanation:

There are three types of systems which refers to universe. They are

1. Open System : In an open system, both energy and matter have external interactions. Example:- boiling water in an open pan.

2. Closed system : In a closed system, only energy has interaction with surroundings.  Example:- boiling water with a lid on the pan.

3. Isolated system : In an isolated system, neither energy-nor matter has external interactions with surroundings. Example : a thermos flask does not exchange energy and matter.

Hence, the correct option is (a) " neither energy nor matter ".

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The right shoe

Explanation:

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Large vehicles need more room to make turns. when you approach an intersection and see a truck ahead signaling a right turn, you
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3 years ago
A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m
kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

Explanation:

a)

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

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

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm

b)

Here we need to find the height of the image first.

This can be done by using the magnification equation:

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

where:

y' is the height of the image

y = 1.75 m is the height of the real person

q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

c)

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

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