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
We have that for the Question "A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that the minimum force that must be applied on the <em>book is</em>
From the question we are told
A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal
Generally the equation for the Force is mathematically given as
F=44N
Therefore
the minimum force that must be applied on the <em>book is</em>
F=44N
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Answer:
Ф = 239.73 rad
Explanation:
α = 12 + 15×t
W = ∫α×dt
= ∫(12 + 5×t)×dt
= 12×t + 2.5×t^2
then:
Ф = ∫W×dt
= ∫(12×t + 2.5×t^2)dt
= 6×t^2 + 5/6×t^3
therefore the angle at t = 4.88s is:
Ф = 6×(4.88)^2 + 5/6×(4.88)^3
= 239.73 rad
Answer:
21.21 m/s
Explanation:
Let KE₁ represent the initial kinetic energy.
Let v₁ represent the initial velocity.
Let KE₂ represent the final kinetic energy.
Let v₂ represent the final velocity.
Next, the data obtained from the question:
Initial velocity (v₁) = 15 m/s
Initial kinetic Energy (KE₁) = E
Final final energy (KE₂) = double the initial kinetic energy = 2E
Final velocity (v₂) =?
Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:
KE = ½mv²
NOTE: Mass (m) = constant (since we are considering the same car)
KE₁/v₁² = KE₂/v₂²
E /15² = 2E/v₂²
E/225 = 2E/v₂²
Cross multiply
E × v₂² = 225 × 2E
E × v₂² = 450E
Divide both side by E
v₂² = 450E /E
v₂² = 450
Take the square root of both side.
v₂ = √450
v₂ = 21.21 m/s
Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.
Idk its really weird because the ridges are making the quarter stand and therefore able to roll
