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Ierofanga [76]
3 years ago
9

Which of the following temparature is approximately equal to room temperature

Physics
1 answer:
Artist 52 [7]3 years ago
6 0

Hello there! :)

\huge\boxed{\text{C. 293K}}

Room temperature is approximately 20°C.

We can automatically eliminate choices B and D since they are not equal to 20°C.

Since some choices use the Kelvin scale, we can convert from Celsius to Kelvin using a simple formula:

K = C° + 273

Find room temperature in degrees <u>Kelvin</u>:

K = 20° + 273

K = 293°

Thus, the correct choice would be <u>C. 293K.</u>

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A gas expands and does PV work on its surroundings equal to 319 J. At the same time, it absorbs 136 J of heat from the surroundi
LiRa [457]

Answer:

The change in energy of the gas during the process is -1.83\times 10^{2} joules.

Explanation:

We can represent this process by the First Law of Thermodynamics, in which gas does work on its surroundings and absorbs heat from there to describe its change in energy. In other words:

Q_{in} - W_{out} = \Delta E

Where:

Q_{in} - Heat absorbed by the gas, measured in joules.

W_{out} - Work done by the gas, measured in joules.

\Delta E - Change in energy, measured in joules.

If we know that Q_{in} = 1.36\times 10^{2}\,J and W_{out} = 3.19\times 10^{2}\,J, the change in energy of the gas is:

\Delta E = 1.36\times 10^{2}\,J-3.19\times 10^{2}\,J

\Delta E = -1.83\times 10^{2}\,J

The change in energy of the gas during the process is -1.83\times 10^{2} joules.

3 0
3 years ago
Many physical quantities are connected by inverse square laws, that is, by power functions of the form f(x)=kx^(-2). In particul
goldenfox [79]

Answer:

  • 4 times

Explanation:

Since the equation for the illumination of an object, i.e. the brightness of the light, is <em>inversely proportional to the square of the distance from the light source</em>, the form of the function is:

  • f(x) = k.x⁻²

Where x is the distance between the object and the light force, k is the constant of proportionality, and f(x) is the brightness.

Then, if you move halfway to the lamp the new distance is x/2 and the new brightness (call if F) is :

F=k(x/2)^{-2}=\frac{k}{(x/2)^2}= \frac{k}{x^2}. 4=f(x).4

Then, you have found that the light is 4 times as bright as it originally was.

3 0
2 years ago
A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.9m/s and the hatchback ca
4vir4ik [10]

Answer:

s = 589.3 m

Explanation:

Let the truck and car meet at a distance = s  m

The truck is moving at constant velocity = v

so s= v * t ---------- (1)

car:

Vi = 0 m/s

a = 3.9 m/s²

s = Vi* t + 1/2 a t²

s= 0 * t +  1/2 a t²

s =  1/2 a t²   ----------- (2)

compare equation (1)  and equation (2)

s= v * t = 1/2 a t²

⇒ v * t = 1/2 a t²

⇒ t = 2 * v/ a

⇒ t = (2 * 33.9 )/ 3.9

⇒ t = 17. 38 s

Now

from equation (1)

s= v * t

s= 33.9 * 17.38

⇒ s = 589.3 m

3 0
3 years ago
A box with a mass of 12.5kg sits on the floor how high would you need to lift it has a GPE of 568j
Degger [83]
GPE=mgh
m= 12.5kg
g= 9.81 always
h=?

568=12.5*9.81*h
Solve for h
You will get 4.63m
4 0
3 years ago
If a sample of an unknown material with a mass of 0.68 g and a volume of 0.8 cm3 is dropped
Anna71 [15]
As the density of the unknown substance is 0.68g/0.8ml = 0.85g/ml, it is less dense than the maple syrup at 1.33g/ml and will float.
4 0
3 years ago
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