In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
I think the answer is c but I’m not sure
Answer:
Friction is when a force is applied or done by weight dragging onto something.
Explanation:
Gravity is when an object is getting pulled toward the center of what is attracting it. And applied force is when someone/sommething is applying force.
Answer:
distance = 33.124 meters
Explanation:
To solve this question, we will use one of the equations of motion which is:
s = ut + 0.5a * t^2
where:
s is the distance that we want to get
u is the initial velocity = 0
a is the acceleration due to gravity = 9.8 m/sec^2
t is the time = 2.6 sec
Substitute with the givens in the equation to get the distance as follows:
s = ut + 0.5a * t^2
s = (0)(2.6) + 0.5(9.8)(2.6)^2
s = 33.124 meters
Hope this helps :)
need speed of sound on lhs