the answer should be:
When the buoyant force is equal to the force of gravity
Answer: 4.7m/s²
Explanation:
According to newton's first law,
Force = mass × acceleration
Since we are given more the one force, we will take the resultant of the two vectors.
Mass = 2.0kg
F1+F2 = (3i-8j)+(5i+3j)
Adding component wise, we have;
F1+F2 = 3i+5i-8j+3j
F1+F2 = 8i-5j
Resultant of the sum of the forces will be;
R² = (8i)²+(-5j)²
Since i.i = j.j = 1
R² = 8²+5²
R² = 64+25
R² = 89
R = √89
R = 9.4N
Our resultant force = 9.4N
Substituting in the formula
F = ma
9.4 = 2a
a = 9.4/2
a = 4.7m/s²
Therefore, magnitude of the acceleration of the particle is 4.7m/s²
It holds A. Less water than when it is warmer.
Solution :
Given :
Mass attached to the spring = 4 kg
Mass dropped = 6 kg
Force constant = 100 N/m
Initial amplitude = 2 m
Therefore,
a). 
= 10 m/s
Final velocity, v at equilibrium position, v = 5 m/s
Now, 
A' = amplitude = 1.4142 m
b). 
m' = 2m
Hence, 
c). 
Therefore, factor
Thus, the energy will change half times as the result of the collision.
