Explanation:
First, find the velocity of the projectile needed to reach a height h when fired straight up.
Given:
Δy = h
v = 0
a = -g
Find: v₀
v² = v₀² + 2aΔy
(0)² = v₀² + 2(-g)(h)
v₀ = √(2gh)
Now find the height reached if the projectile is launched at a 45° angle.
Given:
v₀ = √(2gh) sin 45° = √(2gh) / √2 = √(gh)
v = 0
a = -g
Find: Δy
v² = v₀² + 2aΔy
(0)² = √(gh)² + 2(-g)Δy
2gΔy = gh
Δy = h/2
Answer: acceleration is 6 m/s²
Explanation: V = vo + at. If initial speed Is zero,
V = at and a= V/t = 3m/s /0.5 s
<span>We know that an object in moving with acceleration follow the rule according that
the distance covered will be : d = Vi*t + 1/2*a*t^2
where d is distance, Vi is initial speed, and a is acceleration
Then after 1 km which is 1000 metres we have:
1000 = Vi *71.2 + 1/2*0.0499*(71.2)^2
Vi = (1000-1/2*0.0499*(71.2)^2)/71.2 = 1000/71.2 - 1/2*0.0499*71.2 = 12.27 m/s
Then the car was going at 12.27 m/s when started to accelerate.</span>
Answer:
Option E is correction. None of the above.
Explanation:
( 1 ) A 3p orbital has more energy than 2p orbital and this is the reason it is away from the nucleus as compare to 2p orbital. Energy of the shells increases as their distance increases from the nucleus.
(2) p subshells are made up of three dumbbell-shaped orbitals
(3) There are three atomic orbitals in a p subshell. They are px, py, and pz.