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Troyanec [42]
2 years ago
5

A student left a bar of chocolate in the sun on a hot day. As the chocolate melted, which property changed?

Physics
1 answer:
barxatty [35]2 years ago
7 0
The property changed was its weight

The melted chocolate weighs the same, has an identical mass, and is still composed of chocolate.

Now it’s just melted chocolate instead of a candy bar and has thus changed shapes.
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Two objects, A and B, each of mass o.22 kg, are moving at 0.34 m/s directly toward each other. What is the speed of Object A aft
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Theoretically, if the objects have the same mass and are moving towards each other at a speed of 0.34m s^{-1}, after a perfectly elastic collision, the object A is supposed to move with the same velocity in the opposite direction.
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2 years ago
Why are household wall sockets alternating current (AC) instead of direct current (DC)? Select all that apply.
Bond [772]

Answer:

It is easier to scale the voltage of AC from high to low and low to high than with DC

Explanation:

typically power is used far away from the place where it's generated so to ensure that transmission losses( copper losses) are minimized voltage has to be stepped up during transmission..but due to the fact that most house hold equipment requires low voltage levels it has to be stepped down once it reaches a household/ domestic load...it's easier to do this for Ac than for DC.

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2 years ago
Gravitational Potential Energy = mass x gravity x height
Blizzard [7]
M= gpe / gh
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7 0
3 years ago
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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.

3 0
2 years ago
List three cases in which potential energy becomes
olga55 [171]
1 Electrical Potential Energy, separating two charged plates will store energy as the plates want to return to their original position. 

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<span>3 Gravitational energy is stored by moving something (ball or pendulum are both examples of this) against a gravity gradient (lifting an object) that wants to fall back down. </span>
5 0
3 years ago
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