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 correct answer would be B.) 98°c-102°c
Answer:
if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
Explanation:
The air in the tube can be considered an ideal gas,
P V = nR T
In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H
For pressure the open end of the tube is
P₂ = P_atm + ρ g H
Let's write the gas equation for the colon
P₁ V₁ = P₂ V₂
P_atm V₁ = (P_atm + ρ g H) V₂
V₂ = V₁ P_atm / (P_atm + ρ g h)
If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
The main assumption is that the temperature during the experiment does not change
Answer:
The image is real.
The image is inverted.
The image is bigger than the object.
Explanation:
I really don't know why... I got this question wrong and they said this was the answer. I wish I did. Sorry.
Answer:
A system of two co-axial cylindersof different diameters which rotate together is called wheel and axle example; the door knob , knob of the tap ,screw driver,water tap
