Answer:
Yellow dwarfs are small, main sequence stars. The Sun is a yellow dwarf. A red dwarf is a small, cool, very faint, main sequence star whose surface temperature is under about 4,000 K. Red dwarfs are the most common type of star. Proxima Centauri is a red dwarf.
Explanation:
Answer: 32 centimeters is 12.5984 inches
Explanation:
Answer:
1: 6.637e-13 N
2: 6.637e-09 N
3: 1.335e-08 N
4: 1.335e-08 N
5: 1.456e-06 N
6: 5.839e-07 N
7: 6.673e-11 N
I'd suggest double checking these if you can.
Explanation:
Each of these you can answer by plugging the numbers into the equation
G is the gravitational constant 6.673×10⁻¹¹ N m² / kg²
So the first one would be:
I'm not going to run through showing all calculations for the rest, as you're in a rush, so to whip through them
2: 6.637e-09 N
3: 1.335e-08 N
4: 1.335e-08 N
5: 1.456e-06 N
6: 5.839e-07 N
7: 6.673e-11
Pardon the lack of superscript, I just punched these into a python console to calculate them.
To illustrate clearly, I will rewrite the reaction in a more understandable manner.
<span>2 Al(s) + Fe</span>₂O₃ (s) ⇒ 2 Fe(s) + Al₂O₃(s) Δ<span>hrxn = –850 kJ
This reaction has a negative sign for the change in enthalpy of reaction. The sign convention only means that the reaction releases energy to the surroundings. In other words, the reaction is exothermic. Focusing on only its magnitude, this means that 850 kJ of energy is needed for this reaction of 2 Aluminum moles and 1 mole of </span>Fe₂O₃ to occur.
Now, if you only had an energy of 725 kJ, then the reaction is incomplete but it will still form Iron (Fe). We use stoichiometric calculations as follows:
725 kJ * (2 mol Fe/850 kJ) = 1.7 moles of Fe
Knowing that the molar mass of Fe is 55.6 g/mol, then the mass of produced iron is
1.7 mol Fe * 55.6 g/mol = 94.85 g iron
Answer:
Explanation:
To find the angular velocity of the tank at which the bottom of the tank is exposed
From the information given:
At rest, the initial volume of the tank is:
where;
height h which is the height for the free surface in a rotating tank is expressed as:
at the bottom surface of the tank;
r = 0, h = 0
∴
0 = 0 + C
C = 0
Thus; the free surface height in a rotating tank is:
Now; the volume of the water when the tank is rotating is:
dV = 2π × r × h × dr
Taking the integral on both sides;
replacing the value of h in equation (2); we have:
Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.
Then
Replacing equation (1) and (3)
Finally, the angular velocity of the tank at which the bottom of the tank is exposed = 10.48 rad/s