the formula to calculate the force is, F = MA
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
#SPJ4
Change in velocity of larger moose: (1/3)v - v = -(2/3)v
<span>change in velocity of small moose: (1/3)v - (-v) = (4/3)v </span>
<span>- (change in velocity of larger moose)/(change in velocity of smaller moose) = 2
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>
Answer:
a. A = 0.735 m
b. T = 0.73 s
c. ΔE = 120 J decrease
d. The missing energy has turned into interned energy in the completely inelastic collision
Explanation:
a.
4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax
vmax = 4.0 m/s
¹/₂ * m * v²max = ¹/₂ * k * A²
m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²
A = √1.6 m ² = 1.26 m
At = 2.0 m - 1.26 m = 0.735 m
b.
T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s
T = 2π *√ 10 / 100 *s² = 1.99 s
T = 1.99 s -1.26 s = 0.73 s
c.
E = ¹/₂ * m * v²max =
E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J
E₂ = ¹/₂ * 10 * 4² = 80 J
200 J - 80 J = 120 J decrease
d.
The missing energy has turned into interned energy in the completely inelastic collision
Answer:
The metabolic power for starting flight=134.8W/kg
Explanation:
We are given that
Mass of starling, m=89 g=89/1000=0.089 kg
1 kg=1000 g
Power, P=12 W
Speed, v=11 m/s
We have to find the metabolic power for starting flight.
We know that
Metabolic power for starting flight=
Using the formula
Metabolic power for starting flight=
Metabolic power for starting flight=134.8W/kg
Hence, the metabolic power for starting flight=134.8W/kg