Frost will disturb the smooth flow of air over the wing, unpleasantly
distressing its lifting competence. In other words, this spoils the even flow
of air over the wings, by this means decreasing lifting capability. Also, frost
may avoid the airplane from becoming flying at normal departure speed.
Answer: 6.67 x 10^-11 m^3•kg^-1•s^-2
Answer: q = 2.781e-9C = 2.781nC
E=200C
Explanation:
E = Qd/(2πEor^3)
Where
E=Electric field intensity
Q=Charge
d=distance between the dipole=0.008m
Eo=permitivitty
400 N/C = Q(0.80e-2 m)/(2πε*(10e-2 m)^3)
Q= (400* 2* 3.142 * 8.85 x 10-12 * 0.1^3)/0.008
q = 2.781e-9C = 2.781nC
b)
Though the dipole are two separate charges. And since the point is on the x-axis, the electric field strengths are equivalent. The magnitude of the vector sum is:
E = kq*2sin θ/r^2
= 2(8.99e9 N*m^2/C^2)(2.781e-9 C)*sin(arctan(.4/10))/(10e-2 m)^2
= 2(8.99e9) * (2.781e-9) * sin(2.290)/(10e-2 m)^2
=200 C
Complete Question
A 75 kg man and a 100. Kg woman are playing tug of war on a frictionless surface. The man accelerates toward the woman at 2.0 m/s2.How much force does she use to pull on the rope
Answer:
The value is 
Explanation:
From the question we are told that
The mass of the man is 
The mass of the woman is 
The acceleration of the man is 
Generally the force which she used to pull the rope is mathematically represented as
=> 
=>
Applying the ideal gas equation:
PV = nRT
PV = (mass/Mr)RT
mass = PVMr/RT
mass = (101325 x 4672.2 x 4) / (8.314 x 297)
= 766887.3 kg
= 7.7 x 10⁵ kg
