The center of mass isn't affected by the explosion.
To find the answer, we need to know about the trajectory of motion at zero external force.
<h3>How is the trajectory of an object changed when the net external force on it is zero?</h3>
- When there's no net external force acting on an object, its momentum doesn't change with time.
- As its momentum doesn't change, so it continues with the original trajectory.
<h3>Why doesn't the trajectory of firework change when it's exploded?</h3>
- When a firework is exploded, its internal forces are changed, but there's no external force.
- So, although the fragments follow different trajectories, but the trajectory of center of mass remains unchanged.
Thus, we can conclude that the center of mass isn't affected by the explosion.
Learn more about the trajectory of exploded firework here:
brainly.com/question/17151547
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Answer:
t = 22.2 s
Explanation:
angular distance covered in the 36.0 s is
θ = ω(avg)t = ½(10.0 + 30.0)36 = 720 radians
720/2 = 360 radians
α = Δω/t = (30 - 10)/36 = 5/9 rad/s²
θ = ω₀t + ½αt²
360 = 10.0t + ½(5/9)t²
0 = (5/18)t² + 10.0t - 360
0 = t² + 36t - 1296
t = (-36 ±√(36² - 4(1)(-1296))) / 2
t = (-36 ±√(6480)) / 2
t = -18 ±√1620
we ignore the negative time result as it occurs before we care.
t = -18 + √1620 = 22.249223... s
Answer:
$1.26
Explanation:
Power =energy/ time
energy =powerxtime
energy =50x31x24=37200
=37.2kwh
1kwh =3.39
37.2kwh=3.39x37.2=126.108cent
=$1.26
Answer:
42.96 km/s
Explanation:
From the conservation of Energy
Mass gets cancelled
= Escape velocity of Earth = 11.2 km/s
= Velocity of projectile = 44.4 km/s
The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s
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