Answer:
(a) 27.1 m/s
(b) 3.9 second
Explanation:
Let the speed is u.
Maximum horizontal range, R = 75 m
The range is maximum when the angle of projection is 45°.
(a) Use the formula for the maximum horizontal range


u = 27.1 m/s
(b) Let the time of flight is T.
Use the formula for the time of flight


T = 3.9 second
Answer:
higher, lower, external work, 100 %, work.
Explanation:
These paragraphs refer to the second law of thermodynamics. There are two statements for the second law of thermodynamics. They are as follows:
KELVIN STATEMENT:
All the heat from a source can never be transferred to the sink without the rejection of some heat.
CLAUSIUS STATEMENT:
Heat can not be transferred from a colder body to a hotter body without the application of som external work.
According to the statements the blanks can be filled as follows:
Heat naturally flows from an object that has a <u>higher</u> temperature to an object that has a <u>Lower</u> temperature. Heat can be made to flow in the reverse direction if <u>external work</u> is done. A machine can never have an efficiency of <u>100 %</u>. This means that heat energy can never be fully converted into <u>work</u> energy.
Answer:
Temperature of the hot reservoir is 1540K
Explanation:
![E= 1- \frac{T_{c}}[tex]{T_h}=308+{T_c}\\Efficiency of a carnot engine is given by the aboveTc=temperature of the cold reservoirTh= temperature of the hot reservoirK=273+ 35 (convert 35°C to kelvin)K=308k{T_h}={T_c}+308-----------------------(equation 1)20%=1-{T_c}/{T_h}](https://tex.z-dn.net/?f=E%3D%201-%20%5Cfrac%7BT_%7Bc%7D%7D%5Btex%5D%7BT_h%7D%3D308%2B%7BT_c%7D%5C%5C%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EEfficiency%20of%20a%20carnot%20engine%20is%20given%20by%20the%20above%3C%2Fp%3E%3Cp%3ETc%3Dtemperature%20of%20the%20cold%20reservoir%3C%2Fp%3E%3Cp%3ETh%3D%20temperature%20of%20the%20hot%20reservoir%3C%2Fp%3E%3Cp%3EK%3D273%2B%2035%20%20%28convert%20%2035%C2%B0C%20to%20kelvin%29%3C%2Fp%3E%3Cp%3EK%3D308k%3C%2Fp%3E%3Cp%3E%3Cstrong%3E%7BT_h%7D%3D%7BT_c%7D%2B308-----------------------%28equation%20%201%29%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3E20%25%3D1-%7BT_c%7D%2F%7BT_h%7D)
0.2=({T_c}+308-{T_c})/{T_c}+308
.2({T_c}+61.6=308
0.2{T_c}=246.4
{T_c}=1232
recall from equation 1
{T_h}=308+1232
{T_h}=1540K
Answer:
2 Mg(s) + O2(g) → 2 MgO(s)
As magnesium (Mg) has a valency of +2, and oxygen (O) has a valency of -2, the ratio would be 1:1, and magnesium oxide would be represented as MgO. It is insoluble in water, hence it has the subscript as a solid, represented by (s). In order to use up the diatomic oxygen (O2), there needs to be two moles of magnesium (2Mg) on the reactant side. This would produce 2 moles of MgO on the product side.
Explanation:
<span>Efficiency is the
measure of how efficient a process is. It is used to assess the ability of a
process in avoiding waste energy, materials, money and time in doing a
desirable output. It is calculated as;
Efficiency = useful energy ouput / total energy input</span>
<span>Efficiency = 15(9.81)(20) / 3450 = 0.85</span>