The acceleration of one of those bugs is equal to 305mi/s.
<h3>Acceleration calculation</h3>
To calculate the insect's acceleration, the action and reaction force of the impact must be considered.
As the insect will hit the helmet, the force it hits is the same force it receives, so we can make the following expression:


<em>Speed has been converted to miles per second</em>

So, the acceleration of one of those bugs is equal to 305mi/s.
Learn more about acceleration calculation: brainly.com/question/390784
The velocity of the boy when he hits the water at the bottom of the slide is 14 m/s.
<h3>
Velocity of the boy at the bottom of the slide</h3>
The velocity of the boy when he hits the water at the bottom of the slide is calculated from the principle of conservation of energy.
K.E = P.E
¹/₂mv² = mgh
v² = 2gh
v = √2gh
where;
- h is height of the boy
- g is acceleration due to gravity
v = √(2 x 9.8 x 10)
v = 14 m/s.
Thus, the velocity of the boy when he hits the water at the bottom of the slide is 14 m/s.
Learn more about velocity here: brainly.com/question/6504879
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At 0 Kelvin everything stops moving ( even electrons ), so you could try to decrease the system's temperature
0 Kelvin = -273 Celsius
Answer:
1.40625 kg-m^2
Explanation:
Supposing we have to calculate rotational moment of inertia
Given:
Mass of the ball m= 2.50 kg
Length of the rod, L= 0.78 m
The system rotates in a horizontal circle about the other end of the rod
The constant angular velocity of the system, ω= 5010 rev/min
The rotational inertia of system is equal to rotational inertia of the the ball about other end of the rod because the rod is mass-less

=1.40625 kg-m^2
m= mass of the ball and L= length of the ball
Answer:
the angular displacement Δθ of the tub during a spin of 92.1s is 3122.19 rad or 496.91 rev
Explanation:
Given;
Angular velocity v = 33.9 rad/s
Time t = 92.1 s
Angular displacement d = angular velocity × time
d = vt
Substituting the given values;
d = 33.9 × 92.1 rad
d = 3122.19 rad
To revolutions;
revolution = radian/2π
d = 3122.19/2π rev
d = 496.91 rev
the angular displacement Δθ of the tub during a spin of 92.1s is 3122.19 rad or 496.91 rev