Answer:
The magnitude of the lift force L = 92.12 kN
The required angle is ≅ 16.35°
Explanation:
From the given information:
mass of the airplane = 9010 kg
radius of the airplane R = 9.77 mi
period T = 0.129 hours = (0.129 × 3600) secs
= 464.4 secs
The angular speed can be determined by using the expression:
ω = 2π / T
ω = 2 π/ 464.4
ω = 0.01353 rad/sec
The direction
θ = 16.35°
The magnitude of the lift force L = mg ÷ Cos(θ)
L = (9010 × 9.81) ÷ Cos(16.35)
L = 88388.1 ÷ 0.9596
L = 92109.32 N
L = 92.12 kN
A box moves 5\,\text m5m5, start text, m, end text horizontally when force F=20\,\text NF=20NF, equals, 20, start text, N, end t
Daniel [21]
Answer:
70.71
Explanation:
Given that :
Force (F) = 20N
Distance moved (d) = 5 m
θ = 45°
Workdone on the box :
Workdone (W) = Fdcosθ ; where ;
F = Force, d = Displacement, θ = angle
W = 20 * 5 * cos45°
W = 100 * 0.7071067
W = 70.71 Nm
Answer:
it would have to be 1 or maybe 2
The answer is A because the healthiest option out of all of those selections is the fruit smoothie.
Answer:
Explanation:
We are given that
Charged on alpha particle=q=2e=
Where
Initial kinetic energy=K.E=5.76 MeV=
Z=79
Charge on protons=
We have to find the closeness of alpha particle to the gold nucleus before being turned around.
Initial kinetic energy=Final potential energy
Where