Answer:
713.51 N/m
Explanation:
Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.
From hook's law,
F = ke ...........................Equation 1
Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.
Make k the subject of the equation,
k = F/e ............................ Equation 2
Given: F = 264 N, e = 0.37 m.
Substitute into equation 2
k = 264/0.37
k = 713.51 N/m
Hence the spring constant of the bow = 713.51 N/m
Answer:
1. 
2. 
Explanation:
An ion is formed when an atom that is said to be neutral gains or losses electrons.
It is thought that a negative ion (anion) is produced as it gains electrons and a positive ion (cation) is formed when it loses an electron.
Atomic number is the total number of protons and electrons in a neutral atom.
From the information
Protons = 14
electron = 18
Net Charge = no of proton - no of electron
= 14 - 18 = -4
Mass number = 14 + 15 = 29
Thus, the chemical symbol =
For ion with 27 proton, 32 neutrons and 25 electrons
Net charge = 27 - 25 = +2
Mass number = 27 + 32 = 59
Thus, the chemical symbol =
Answer:
5×10⁵ L of ammonia (NH3)
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N2 + 3H2 —> 2NH3
From the balanced equation above, we can say that:
3 L of H2 reacted to produce 2 L of NH3.
Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:
From the balanced equation above,
3 L of H2 reacted to produce 2 L of NH3.
Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.
Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.
