Answer:
The velocity of the first block is 1.15m/s while of the second block 2.56m/s.
Explanation:
Momentum is only conserved in an isolated system, and because this problem requires us to find the value of the two variables, we need two equations; therefore, to conserved momentum the energy must be released in to the system only after the collision has occurred.
Therefore, from conservation of momentum



and from conservation of energy



Thus, we have two equations and two unknowns


which has solutions
and

Since the blocks cannot pass through each other, the 0.5kg block cannot have
(moves to the left) while the 0.4 kg block has
(moves to the right); therefore, we take the first solution for the velocities:
.
Thus , the velocity of the first block is 1.15m/s while of the second block 2.56m/s.
The approximate acceleration of gravity for an object above the earth's surface is 9.8 m/s².
<h3>
Acceleration due to gravity </h3>
The acceleration due to gravity of an object above the Earth's surface is calculated from Newton's second law of motion and Newton's law of universal gravitation.
<h3>Force between the object and Earth</h3>

<h3>Weight of the object</h3>

Solve (1) and (2)
mg = GmM/R²
g = GM/R²
where;
- M is mass of Earth
- G is gravitational constant
- R is radius of the Earth = 6,371 km
g = (6.673 x 10⁻¹¹ x 5.98 x 10²⁴)/(6,371 x 10³)²
g = 9.8 m/s²
Learn more about acceleration due to gravity here: brainly.com/question/88039
#SPJ1
For Plato, the correct answer is Concave mirror :)
Answer:
Option C, The total momentum of the fragments is equal to the original momentum of the firecracker.
Explanation:
Kinetic energy of cracker cannot remain constant before and after explosion. It is so because in the process of burning and bursting some amount of kinetic energy is lost in the form of light and heat energy. While the total mass before and after the explosion remains constant due to which the momentum is conserved before and after the explosion
Hence, option C is correct
Explanation:
a) Given in the y direction (taking down to be positive):
Δy = 50 m
v₀ = 0 m/s
a = 10 m/s²
Find: t
Δy = v₀ t + ½ at²
50 m = (0 m/s) t + ½ (10 m/s²) t²
t = 3.2 s
b) Given in the x direction:
v₀ = 12 m/s
a = 0 m/s²
t = 3.2 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (12 m/s) (3.2 s) + ½ (0 m/s²) (3.2 s)²
Δx = 38 m