Answer:
The terminal velocity of the diver is 115 m/s = 414 km/hr
Explanation:
At terminal velocity,
Fnet = mg - Fd = 0
Drag force, Fd = cρAv²/2
mg = cρAv²/2
Terminal Velocity of a body falling through a fluid as in a diver falling through air is given by
v = √(2mg/ρcA)
where m = mass of body falling through fluid = 80 kg
g = acceleration due to gravity = 9.8 m/s²
ρ = density fluid, density of air, as obtained from literature = 1.21 kg/m³
c = coefficient of drag friction of diver falling through air, as obtained from literature = 0.7
A = the area of the diver facing the fluid = 0.14 m²
v = √(2mg/ρcA) = √((2 × 80 × 9.8)/(1.21 × 0.7 × 0.14)) = 115 m/s = 115 × (3600/1000) km/hr = 414 km/hr
The solution would be like this for this specific problem:
Given:
diffraction grating
slits = 900 slits per centimeter
interference pattern that
is observed on a screen from the grating = 2.38m
maxima for two different
wavelengths = 3.40mm
slit separation .. d =
1/900cm = 1.11^-3cm = 1.111^-5 m <span>
Whenas n = 1, maxima (grating equation) sinθ = λ/d
Grant distance of each maxima from centre = y ..
<span>As sinθ ≈ y/D y/D =
λ/d λ = yd / D </span>
∆λ = (λ2 - λ1) = y2.d/D - y1.d/D
∆λ = (d/D) [y2 -y1]
<span>∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m</span></span>
Answer:
E = 2k 
Explanation:
Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.
We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.
Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀
tells us that the linear charge density is
λ = q_ {int} /l
q_ {int} = l λ
we substitute
E A = l λ /ε₀
is area of cylinder is
A = 2π r l
we substitute
E =
E =
the amount
k = 1 / 4πε₀
E = 2k 
Answer:
<h3>62.5N</h3>
Explanation:
The pressure at one end of the piston is equal to the pressure on the second piston.
Pressure = Force/Area
F1/A1 = F2/A2
Given
F1 = 250N
A1 = 2.0m²
A2 = 0.5m²
F2 = ?
Substituting the given values in the formula;
250/2 = F2/0.5
cross multiply
250*0.5 = 2F2
125 = 2F2
F2 = 125/2
F2 = 62.5N
Hence the force needed to lift this piston if the area of the second piston is 0.5 m^2 is 62.5N