A)After 3,5s -->v=v0+gt=21+(-9,8•3,5)=21+(-34,3)=-13,3m/s;
b)The maximum height that the orange reaches is h max=v0^2/2g=22,5m;v^2=sqrt(2gh)=>h=v^2/2g=9,025m.The height of the orange is H=h max-h=13,475m.;
c)The orange is traveling down.
The frequencies of light that an atom can emit are dependent on states the electrons can be in. When excited, an electron moves to a higher energy level or orbital. When the electron falls back to its ground level the light is emitted.
hope this helped:)
mark brainliest
The answer is false, your welcome.
Question:
A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?
Answer:
Time for the race will be t = 9.26 s
Explanation:
Given data:
As the sprinter starts the race so initial velocity = v₁ = 0
Distance = s₁ = 20 m
Acceleration = a = 4.20 ms⁻²
Distance = s₂ = 100 m
We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.
Using 3rd equation of motion
(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)
v₂ = 12.96 ms⁻¹
Time for 20 m distance = t₁ = (v₂ - v ₁)/a
t₁ = 12.96/4.2 = 3.09 s
He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be
Time for 100 m distance = t₂ = s₂/v₂
t₂ = 80/12.96 = 6.17 s
Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s
T = 9.26 s
We actually cant, really. Black Holes are really very powerful and don't just happen in some random place in the cosmos. We have got shots of Black holes that you cannot even see, but all the other big black holes that look like something straight of a sci-fi movie that look real, are really just photo- shopped. good question. Hope This Helped.