Ox:vₓ=v₀
x=v₀t
Oy:y=h-gt²/2
|vy|=gt
tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g
y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°
cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s
Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ
Inertia
the awnswer is inertia b
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:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
