Answer:
Explanation:
Potential energy is the energy of a body due to is virtue of rest.
Potential energy is given as mgh
g is a constant and it is 9.81m/s²
And also the mass of the body is given as 1.3kg
Now the height of the body is
He took a book to a storey building of height 26m
He still holds the book 151 cm (1.51m) above the house.
The house is on an altitude of 1609m from the sea level.
Total Ug with out the sea level is
Ug=mgh
Ug=1.3 × 9.81 ×(26+1.51)
Ug=350.84J
Then, the potential energy due to the sea level is given as
Ug=mgh
Where g = 1/6371 m/s²
Therefore
Ug=mgh
Ug=1.3 × 1/6371 ×1609
Ug=0.328J
Total energy = 0.328+350.84
Ug=351.17J
It stays constant, because it's using that energy to change state
Answer:
L/D= 112
Explanation:
Aerodynamics can be defined as the branch of dynamics which deals with the motion of air, their properties and the interaction between the air and solid bodies.
Aerodynamics law explains how an airplane is able to fly. There are four forces of flight, and they are; lift, weight, thrust and drag. The amount of lift generated by a wing divided by the aerodynamic drag is known as the lift to drag ratio.
Lift increases proportionally to the square of the speed.
The solutions to the question is the file attached to this explanation.
Lift,L= qC(l). S---------------------------(1).
and,
Drag,D = qC(d).S ----------------------(2).
Hence, Lift to drag ratio,L/D= C(l)/C(d).
Therefore, we have to compute various angle of attack.(check attached file)...
Then, (L/D) will then be equal to 112.
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s