Answer:
D)evaluating a solution
Explanation:
In this scenario, the next logical step would be evaluating a solution. This is because Jasper and Samantha have already identified the problem/need which is that the robot needs to be able to move a 10-gram weight at least 2 meters and turn in a circle. They also designed and implemented a solution because they have already built the robot. Therefore the only step missing is to evaluate and make sure that the robot they built is able to complete the requirements.
Answer:
m = 35.98 Kg ≈ 36 Kg
Explanation:
I₀ = 125 kg·m²
R₁ = 1.50 m
ωi = 0.600 rad/s
R₂ = 0.905 m
ωf = 0.800 rad/s
m = ?
We can apply The law of conservation of angular momentum as follows:
Linitial = Lfinal
⇒ Ii*ωi = If*ωf <em>(I)</em>
where
Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m
If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m
Now, we using the equation <em>(I) </em>we have
(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800
⇒ m = 35.98 Kg ≈ 36 Kg
Well you’d have a force due to gravity, the normal force which will be perpendicular to the sources (meaning you’ll have components to this vector), and you’d have the force of friction opposing the motion of the box. I’m also assuming there’s no air resistance. In this case you’d have three vector forces.
Atoms is basic particles ,electrons,neutrons and the Regions of the atom are called electron shells and contain the electrons. So “a neutral core surrounded by mostly empty space”sounds pretty sure to me :)
Answer:
-6327.45 Joules
650.375 Joules
378.47166 N
Explanation:
h = Height the bear slides from = 15 m
m = Mass of bear = 43 kg
g = Acceleration due to gravity = 9.81 m/s²
v = Velocity of bear = 5.5 m/s
f = Frictional force
Potential energy is given by
Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules
Kinetic energy is given by
Kinetic energy of the bear just before hitting the ground is 650.375 Joules
Change in total energy is given by
The frictional force that acts on the sliding bear is 378.47166 N
