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
Answer:
W = 0 J
Explanation:
The amount of work done by gas at constant pressure is given by the following formula:
where,
W = Work done by the gas
P = Pressure of the gas
ΔV = Change in the volume of the gas
Since the volume of the gas is constant. Therefore, there is no change in the volume of the gas:
<u>W = 0 J</u>
m = 5 kg
a = 2 m/s²
to find the force that accelerates the 4 kg object @ 2 m/s²
F = ma = 5 kg x 2 m/s² = 10 N
To find what acceleration 10 N would give a 20 kg object
a = F/m = 10 N/20 kg = 0.5 m/s
Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
Answer:
a) force between them is attraction, b) F = 1.125 10⁻² N
Explanation:
In this case the electric force is given by Coulomb's law
F =
In the exercise they give us the values of the loads
q1 = - 10 mC = -10 10⁻³ C
q2 = 5 mC = 5 10⁻³ C
d = 20 cm = 0.20 m
let's calculate
F = 9 10⁹ 
F = 1.125 10⁻² N
To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction
