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NeX [460]
3 years ago
8

Fon of the assignment will be graded by your teacher Question: In a lab experiment, A square of black paper and a square of whit

e paper are place directly under 100-watt lamps for 30 minutes. Predict what happens to the temperature of each paper during the experiment.​
Physics
1 answer:
ser-zykov [4K]3 years ago
6 0

Answer: OVERVIEW

Students will use black and white construction paper and a light source to learn that dark objects

absorb more light and reflect less light than bright objects. The activity also demonstrates the conversion

of radiant light energy into heat energy.

CONCEPTS

• Dark surfaces absorb more visible light energy than bright surfaces

• Dark surfaces reflect less visible light energy than bright surfaces

• Energy can change forms, in this case from radiant light energy to heat

• Clouds, being bright, reflect significant amounts of sunlight and help to regulate Earth’s temperature

MATERIALS

• 2 thermometers

• Flood lamp, desk lamp, or area in direct sunlight

• Ruler

• Construction paper, 1 piece white, 1 piece black, or 2 sheets photocopy paper

• Scissors

• Cellophane tape or rubber bands

• 2 empty metal food cans, same size (be sure rims are not jagged)

PREPARATION

The paper and the cans can be prepared beforehand or prepared as part of the activity (see Procedure).

Although two cans with their tops completely removed can be used, the experiment will be more effective

(have fewer external effects), if only holes are placed in the cans’ lids, e.g., two holes from a bottle opener

to empty material out of the can, and one center hole created with an awl for the thermometer. Only one

hole is actually needed for the experiment - for the thermometer. Cans can optionally be filled with water,

or this can be done as a separate experiment to demonstrate the higher heat capacity of water compared to

air.

You can either use a flood lamp or a desk lamp (light bulb) to simulate sunlight, as described here, or

you can place the cans on a windowsill (window closed) or other sheltered area in direct sunlight. A flood

lamp will be the most effective option, causing the largest temperature increases.

PROCEDURE

Engagement

Discuss whether dark surfaces (e.g., asphalt) or bright surfaces (e.g., concrete) tend to get hotter in

sunlight. Which would you rather walk on during the day in the summertime? What color are solar cells,

for example those found on some calculators or freeway call boxes?

Explanation: I hope this helps! ∧    ∧

                                                ⊂∵→ω←∵⊃

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So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th
tia_tia [17]

Answer:

a) a=33.73mm/s^{2}

b) mg>N

c) \%_{change}=0.343\%

d) a=24.07mm/s^{2}

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

a_{c}=\omega ^{2}r

where:

\omega=\frac{2\pi}{T}

we know the period of rotation of the earth is about 24 hours, so:

T=24hr*\frac{3600s}{1hr}=86400s

so we can now find the angular speed:

\omega=\frac{2\pi}{86400s}

\omega=72.72x10^{-6} rad/s^{2}

So the centripetal acceleration will be:

a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)

which yields:

a_{c}=33.73mm/s^{2}

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

\sum F=0

so we get that:

N-mg+ma_{c} = 0

and solve for the normal force:

N=mg-ma_{c}

In this case, we can clearly see that:

mg>mg-ma_{c}

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

mg=(60kg)(9.81m/s^{2})=588.6N

and let's calculate the normal force:

N=m(g-a_{c})

N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})

N=586.58N

so now we can calculate the percentage change:

\%_{change} = \frac{mg-N}{mg}x100\%

so we get:

\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%

\%_{change}=0.343\%

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

cos \theta = \frac{AS}{h}

In this case:

cos \theta = \frac{r}{R_{E}}

so we can solve for r, so we get:

r= R_{E}cos \theta

in this case we'll use the average radius of earch which is 6,371 km, so we get:

r = (6371x10^{3}m)cos (44.4^{o})

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

a=\omega ^{2}r

a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m

a=24.07 mm/s^{2}

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