Answer:
The velocity of block B when the energy stored in the spring bumpers is max = 2.5m/s
Explanation:
An head on collision is aka elastic collision and in elastic collision momentum and energy is conserved that is the total momentum before collision = total momentum after collision
Given
Mass of block A mA = 3kg, initial Velocity of block A vA1 = 5m/s, Final velocity = vA2
Mass of block B mB = 9kg, Initial velocity of block B vB1 = 0m/s, Final velocity = vB2
Therefore we have the equation
1)mAvA1 + mBvB1 = mAvA2 + mBvB2
Also in elastic collision between 2 objects, the relative velocities before and after the collision have the same magnitude but opposite direction
vA1 - vB1 = vB2 - vA2
vA2 = vB2 - vA1 when vB1 = 0
Substitute into the equation we have
(3*5)+(9*0) = 3*(vB2-vA1) + (9*vB2)
15 = 3vB2 - 3vA1 +9vB2
15=3vB2 - (3*5) + 9vB2
15+15 = 12vB2
vB2 = 2.5m/s
Answer:
Answer is 98 joules
Explanation:
P. E=2*9.8*3=98
S.i unit of potential energy ⚡ is joules
So answer is 98 joules
"Confounding variable" is the choice among the following choices given in the question that best describes two <span>or more explanatory variables that are not separated. The correct option among all the options that are given in the question is the second option or option "b". I hope the answer has come to your help.</span>
Answer: its speed upon release is 26.05 m/s
Explanation:
Given that;
mass m = 0.244 kg
force F = 30.3 N
V1 = 14.7 m/s
r = 59.3 cm = 0.593 m
Vf = ?
we know that;
1/2mV1² + FπR = 1/2mVf²
so we substitute
[1/2×0.244×(14.7)²] + [30.3×π×0.593 = 1/2×0.244×Vf²
26.3629 + 56.4478 = 0.122Vf²
82.8107 = 0.122Vf²
Vf² = 82.8107 / 0.122Vf
Vf² = 678.7762
Vf = √678.7762
Vf = 26.05 m/s
Therefore its speed upon release is 26.05 m/s
Friction is caused by two solids rubbing together quickly.