<span>Ans : Initial E = KE = ½mv² = ½ * 1.2kg * (2.2m/s)² = 2.9 J
max spring compression where both velocities are the same: conserve momentum:
1.2kg * 2.2m/s = (1.2 + 3.2)kg * v → v = 0.6 m/s
which means the combined KE = ½ * (1.2 + 3.2)kg * (0.6m/s)² = 0.79 J
The remaining energy went into the spring:
U = (2.9 - 0.79) J = 2.1 J = ½kx² = ½ * 554N/m * x²
x = 0.0076 m ↠(a)</span>
The so-called "terminal velocity" is the fastest that something can fall
through a fluid. Even though there's a constant force pulling it through,
the friction or resistance of plowing through the surrounding substance
gets bigger as the speed grows, so there's some speed where the resistance
is equal to the pulling force, and then the falling object can't go any faster.
A few examples:
-- the terminal velocity of a sky-diver falling through air,
-- the terminal velocity of a pecan falling through honey,
-- the terminal velocity of a stone falling through water.
It's not possible to say that "the terminal velocity is ----- miles per hour".
If any of these things changes, then the terminal velocity changes too:
-- weight of the falling object
-- shape of the object
-- surface texture (smoothness) of the object
-- density of the surrounding fluid
-- viscosity of the surrounding fluid .
Take 68.2/60 = 1.137 hr
take 56.9/1.137 = 50.043 mi/hr
take 189/211 = 0.896
24.8/2 = 12.4 m
12.4/82.3 = 0.15s
Answer:
V is approximately = 23m/s
Explanation:
Kinetic energy = ½ mv²
Where m= mass = 0.450kg
V= velocity =?
K. E = 119J
Therefore
K. E = ½ mv²
Input values given
119= ½ × 0.450 × v²
Multiply both sides by 2
119 ×2 = 2 × 1/2 × 0.450 × v²
238= 0.450v²
Divide both sides by 0.450
238/0.450 = 0.450v²/0.450
v² = 528.89
Square root both sides
Sq rt v² = sq rt 528.89
V = 22.998m/s
V is approximately = 23m/s
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