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3 years ago

13

A 32-ounce stack of paper is at a temperature of 65°F. Then the paper is divided into fourths. The amount of thermal energy in e

ach part is about the amount of thermal energy in the original 32-ounce stack of paper.
A. The Same As

B. Four Times

C. One-Fourth

1 answer:

Yakvenalex [24]3 years ago

8 0

C because the stack of paper was divided into 4

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3. What is the frequency of a wave that has a wave speed of 20 m/s and a wavelength of 0.50 m?

bonufazy [111]

Explanation:

everything can be found in the picture

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3 years ago

An observer that moves with an electrical charge senses:

Shtirlitz [24]

Here observer is moving along with an electrical charge

now magnetic field on a moving charge is given by equation

$F = qvB sin\theta$

electrical force on a moving charge is given by formula

$F = qE$

so here we can say when observer is moving with the charge it will have two forces on it

1. magnetic force due to existing magnetic field

2. electric force due to electric field

so here he must have to experience both type of field presence

So here correct answer must be

<em>C- electric and magnetic fields </em>

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3 years ago

Read 2 more answers

How does the intensity of a sound wave change if the distance from the

Alex17521 [72]

Answer:

The new intensity decreases by a factor of 16.

Explanation:

The intensity of sound wave is given by :

$I=\dfrac{P}{A}$

P is power

A is area

$I=\dfrac{P}{4\pi r^2}$

or

$I\propto \dfrac{1}{r^2}$ , r is distance from the source

If the distance from the source is increased by a factor of 4, r' = 4r

So,

$I'=\dfrac{1}{r'^2}\\\\I'=\dfrac{1}{(4r)^2}\\\\I'=\dfrac{1}{16}\times \dfrac{1}{r^2}\\\\I'=\dfrac{I}{16}$

So, the new intensity decreases by a factor of 16.

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3 years ago

The focal length of the lens of a simple camera is 40.0 mm. how far must the lens be moved to change the focus of the camera fro

Minchanka [31]

.05 cm far must the lens be moved to change the focus of the camera from a person 25 m away to one that is 4.0 m away

The focal length of a thin lens in air is the separation between its major foci, also known as focal points, and its center. The focal length is the distance at which a beam of collimated light will be concentrated to a single spot for a converging lens (for instance, a convex lens).

If v is the image distance in the first scenario

1 / v - 1 / -25 = 1 / .05

1 / v = 1 / .05 - 1 / 25

= 20 - .04 = 19.96

v = .0501 m = 5.01 cm

The second instance

u = 4 ,

1 / v - 1 / - 4 = 1 / .05

1 / v = 20 - 1 / 4 = 19.75

v = .0506 = 5.06 cm

Therefore, the lens must be advanced by 5.06 - 5.01 =.05 cm.

Learn more about lens here brainly.com/question/17086993

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1 year ago

Henry ran a 100 m race. Use the graph to answer the following:

horrorfan [7]

Using the graph, which describes how Henry ran the 100m race;

a) It takes Henry 20seconds to run 100m

b) Henry's average speed over the race is; 5m/s.

According to the linear graph which describes the distance ran by Henry during the 100m race as a function of time.

a) Since the distance from start ran by Henry is plotted on the vertical axis, and the time is plotted on the horizontal axis;

To determine how long it took Henry to run 100m; The point corresponding to 100m is traced downward from the line of the graph and we find out that;

It takes Henry 20seconds to run 100m

b) Henry's average speed over the race is simply;

The slope of the distance-time graph.

Therefore,

Average speed = (100-0)/(20-0)

Average speed = 100/20

Average speed = 5m/s.

Therefore, Henry's average speed over the race is; 5m/s.

Read more:

brainly.com/question/22125199

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2 years ago