Answer:
New volume (V2) = 60 liter
Explanation:
Given:
Amount of helium (V1) = 20 Liter
Temperature (T1) = 100°K
Temperature (T2) = 300°K
Find:
New volume (V2)
Computation:
According to Gas Law:
V1 / T1 = V2 / T2
20 / 100°K = V2 / 300°K
V2 = 60 liter
New volume (V2) = 60 liter
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Answer: Earth's orbital path around the Sun</h2><h2>
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The <u>Ecliptic</u> refers to the orbit of the Earth around the Sun. Therefore, <u>for an observer on Earth it will be the apparent path of the Sun in the sky during the year, with respect to the "immobile background" of the other stars.</u>
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It should be noted that the ecliptic plane (which is the same orbital plane of the Earth in its translation movement) is tilted with respect to the equator of the planet about
approximately. This is due to the inclination of the Earth's axis.
Hence, the correct option is Earth's orbital path around the Sun.
Answer:
The unknown substance is Aluminum.
Explanation:
We'll begin by calculating the change in the temperature of substance. This can be obtained as follow:
Initial temperature (T₁) = 25 ⁰C
Final temperature (T₂) = 100 ⁰C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 100 – 25
ΔT = 75 ⁰C
Finally, we shall determine the specific heat capacity of the substance. This can be obtained as follow:
Change in temperature (ΔT) = 75 ⁰C
Mass of the substance (M) = 135 g
Heat (Q) gained = 9133 J
Specific heat capacity (C) of substance =?
Q = MCΔT
9133 = 135 × C × 75
9133 = 10125 × C
Divide both side by 10125
C = 9133 / 10125
C = 0.902 J/gºC
Thus, the specific heat capacity of substance is 0.902 J/gºC
Comparing the specific heat capacity (i.e 0.902 J/gºC) of substance to those given in the table above, we can see clearly that the unknown substance is aluminum.
The strength of the gravitational field is given by:

where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:

And now we can use the previous equation to calculate the field strength at that altitude:

And we can see this value is a bit less than the gravitational strength at the surface, which is

.