Answer:
A jet pilot puts an aircraft with a constant speed into a vertical loop is explained below in complete details.
Explanation:
Well, the difficulty does not provide the pilot's mass (or weight in regular gravity), but the difficulty can be resolved and declared in courses of m (the pilot's mass).
When the jet is at the foundation of the circuit, a free-body chart displays the centripetal energy working upward approaching the middle of the loop, and the sound force of the chair and the pilot also upward. The pilot's weight (mg) is earthward. From Newton's second law:
?F(c) = ma(c) = n - mg
n = mg + ma(c)
= m[g + a(c)]
Since centripetal acceleration equals v² / r, the equalization enhances:
n = m[g + (v² / r)]
Answer:
E(0.1m)=-16.53.10^6 V.m
E(0.465m)=-3.55.10^6 V.m
E(1.3m)=-1.27^6 V.m
Explanation:
You can find the field using Gauss's Law:
the surface S is an "infinite long" cylinderr of radio r.
E(r)=
λ=-92.0 μC/m, ε=8.85.10^-12
E(0.1m)=-16.53.10^6 V.m
E(0.465m)=-3.55.10^6 V.m
E(1.3m)=-1.27^6 V.m
4. 1 and 2 only.
1. the downward force is the force of gravity.
<span>2. The upward force exerted is the Normal reaction from the floor.</span>
Givens:
Acceleration (a) = 32.17 ft/s² as that is the only force acting on the stone in free fall assuming no air resistance
Initial velocity (V1) = 0 ft/s as that it the starting speed of the stone as all objects about to move from rest must start at 0 velocity at rest = 0 velocity
Final velocity (V2) = 120 ft/s
Displacement aka. the
height of the cliff (Δd) = ? ft
The kinematic formula we can use is: V2² = V1² + 2(a)(Δd)
Isolate to solve for Δd and just plug in the numbers!
V2² = V1² + 2(a)(Δd)
(120)² = (0)² + 2(32.17)(Δd)
14400 = 2(32.17)(Δd)
= Δd
223.9 = Δd
Therefore, the height of the cliff is 223.9 ft! Hope this helped! :)
Answer:
astatine
Explanation:
Sorry if this is wrong i am not a hundred percent sure.