Answer:
(B) The balls reach the ground at the same instant.
Explanation:
Since it is a projectile motion, the time required to reach the ground is given by
t = √2h/g
Where h is the height from which the balls are thrown and g is the acceleration due to gravity.
Since both of the balls were thrown from the same height and the acceleration of both balls are also same they will reach the ground at the same instant.
Answer:
The bond energy of F–F = 429 kJ/mol
Explanation:
Given:
The bond energy of H–H = 432 kJ/mol
The bond energy of H–F = 565 kJ/mol
The bond energy of F–F = ?
Given that the standard enthalpy of the reaction:
<u>H₂ (g) + F₂ (g) ⇒ 2HF (g)</u>
ΔH = –269 kJ/mol
So,
<u>ΔH = Bond energy of reactants - Bond energy of products.</u>
<u>–269 kJ/mol = [1. (H–H) + 1. (F–F)] - [2. (H–F)]</u>
Applying the values as:
–269 kJ/mol = [1. (432 kJ/mol) + 1. (F–F)] - [2. (565 kJ/mol)]
Solving for , The bond energy of F–F , we get:
<u>The bond energy of F–F = 429 kJ/mol</u>
Explanation:
Given that,
Mass of the block, m = 5 kg
Spring constant, k = 2000 N/m
The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back towards equilibrium.
(a) The angular frequency is given by :
Since, , f is frequency
(b) The velocity of particle executing SHM is given by :
x is displacement from equilibrium position
(c) The total mechanical energy of the motion in SHM is given by :
Hence, this is the required solution.