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Ghella [55]
3 years ago
12

A student gathered two boxes of the same size made of different materials: glass and clear plastic. She placed them on a windows

ill in the sun for an hour, and then measured the temperature of the air in each box. What is the dependent variable
Physics
1 answer:
Aleks04 [339]3 years ago
4 0
The temperature in the boxes
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Our Sun is considered a(n) _____ star.<br> A. small <br> B. large <br> C. giant <br> D. average
faltersainse [42]
Checking a table of star sizes, we'll see that our Sun is a medium sized star, so the answer is D, average.
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3 years ago
Is there more than one universe that can be explained?
Firlakuza [10]

No.  We don't know of the existence of another universe besides
the one we live in, and we can't even explain THIS one.

7 0
4 years ago
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A balloon has a volume of 3.5-L at 25^ * C . What would be the volume of the balloon if it were placed in a container of hot wat
Romashka-Z-Leto [24]

Answer:

4.3 L

Explanation:

Ideal gas law:

PV = nRT

Rearrange:

V / T = nR / P

Since n, R, and P are constant:

V₁ / T₁ = V₂ / T₂

Plug in values and solve:

(3.5 L) / (25 + 273.15 K) = V / (95 + 273.15 K)

V = 4.3 L

5 0
3 years ago
We want to find how much charge is on the electrons in a nickel coin. Follow this method. A nickel coin has a mass of about 5.7
Aleksandr-060686 [28]
<span>5.6x10e22 atoms
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8 0
3 years ago
In this problem you will consider the balance of thermal energy radiated and absorbed by a person.Assume that the person is wear
Viefleur [7K]

Answer:P=14.6 W

Explanation:

According to the Stefan-Boltzmann law for real radiating bodies:

P=\sigma A \epsilon T^{4} (1)

Where:

P is the energy radiated (in Watts)

\sigma=5.67(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A is the Surface area of the body  

T=30\°C + 273.15= 303.15 K is the effective temperature of the body (its surface absolute temperature) in Kelvin

\epsilon=0.6 is the body's emissivity

On the other hand, we are told the human body is roughly approximated to a cylinder of length L=2.0m and circumference C=0.8m.

The circumference of a circle is:C=0.8m=2 \pi r where r is the radius. Hence r=\frac{0.8m}{2 \pi}=0.1273 m.

Now we have to input this value for r  in the Area of a cylinder formula:

A=\pi r^{2}L

A=\pi (0.1273 m)^{2}(2 m)

A=0.0509 m^{2} (2)

Substituting (2) in (1):

P=(5.67(10)^{-8}\frac{W}{m^{2} K^{4}}) (0.0509 m^{2}) (0.6) (303.15 K)^{4} (3)

Finally:

P=14.62 W \approx 14.6 W

5 0
3 years ago
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