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.
Answer:
t = 1.75
t = 0.04
Explanation:
a)
For part 1 we want to use a kenamatic equation with constant acceleration:
X = 1/2*a*t^2
isolate time
t = sqrt(2X / a)
Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2
t = sqrt(2*15m / 9.8m/s^2)
t = 1.75 s
b)
The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:
v = d / t
isolate time
t = d / v
plug in known variables
t = 15m / 340m/s
t = 0.04 s
The answer is B friction force
Answer:
R= 602 .11 N
Explanation:
The horizontal component of tension T will give reaction of the wall and the vertical component of T will balance the weight of of the climber .
T cos32 = R
710 x .848 = R
R= 602 .11 N .
Answer:
The answer is 15000kgms^-2
Explanation:
Actually here we are using the formula F=ma
