Answer:
They break down rocks.
Explanation:
The rain is slightly acidic due to the presence of carbon dioxide. Power plants and automobiles produce gases such as sulfur dioxide and nitric oxide, which react with water vapor in the atmosphere to form acid rain. They help in weathering the rocks which makes up the Earth's crust.
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Answer;
The temperature change for the second pan will be lower compared to the temperature change of the first pan
Explanation;
-The quantity of heat is given by multiplying mass by specific heat and by temperature change.
That is; Q = mcΔT
This means; the quantity of heat depends on the mass, specific heat capacity of a substance and also the change in temperature.
-Maintaining the same quantity of heat, with another pan of the same mass and greater specific heat capacity would mean that the change in temperature would be much less lower.
Answer:
Ro = 133 [kg/m³]
Explanation:
In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.
where:
m = mass [kg]
V = volume [m³]
We will convert the units of length to meters and the mass to kilograms.
L = 15 [cm] = 0.15 [m]
t = 2 [mm] = 0.002 [m]
w = 10 [cm] = 0.1 [m]
Now we can find the volume.
![V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ]](https://tex.z-dn.net/?f=V%20%3D%200.15%2A0.002%2A0.1%5C%5CV%20%3D%200.00003%20%5Bm%5E%7B3%7D%20%5D)
And the mass m = 4 [gramm] = 0.004 [kg]
![Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}]](https://tex.z-dn.net/?f=Ro%20%3D%200.004%2F0.00003%5C%5CRo%20%3D%20133%20%5Bkg%2Fm%5E%7B3%7D%5D)
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
