Answer:
The football leaves with the velocity, u = 15.68 m/s
Explanation:
Given data,
The football bounces back up off the ground and is airborne for, t = 3.2 s
Let the football bounces back up off the ground in the vertical direction
The formula for time of flight is given by,
t = 2u /g
∴ u = gt / 2
Substituting the values,
u = 9.8 x 3.2 / 2
u = 15.68 m/s
Hence, the football leaves with the velocity, u = 15.68 m/s
Answer:
The correctly equation is D) N2+ 3H2 → 2NH3
Explanation:
Both nitrogen and hydrogen are presented in bimolecular form (N2 and H2), on one side and on the other side of the equation we have 2 Nitrogens and 6 Hydrogens (it is balanced)
Momentum before the hit:
p = mv = 0.01 * 300 + 1 * 0
Momentum after the hit:
p = 0.01 * 150 + 1 * v
Momentum is conserved:
0.01 * 300 = 0.01 * 150 + v
3 = 1.5 + v
v = 1.5
The velocity of the block after the collision is 1.5 m/s.
Explanation:
By the second law of Newton we get the relation
F = ma
