Answer:
The approximate terminal velocity of a sky diver before the parachute opens is 320 km/h.
Explanation:
- The terminal velocity is the maximum magnitude of velocity that is attained by the diver when he or she falls in the air.
- The terminal velocity of the person diving in air before opening parachute is 320 km/h that means the velocity when the person is experiencing free fall is 320 km/h.
- During terminal velocity, we can represent mathematical equation as;
Buoyancy force + drag force = Gravity
Detailed Explanation:
1) Rusting of Iron
4Fe + 3O2 + 2H2O -> 2Fe2O32H2O
Reactants :-
Fe = 4
O = 3 * 2 + 2 = 8
H = 2 * 2 = 4
Products :-
Fe = 2 * 2 = 4
O = 2 * 3 + 2 = 8
H = 2 * 2 = 4
2) Fermentation of sucrose…
C12H22O11 + H2O -> 4C2H5OH + 4CO2
Reactants :-
C = 12
H = 22 + 2 = 24
O = 11 + 1 = 12
Products :-
C = 4 * 2 + 4 = 12
H = 4 * 5 + 4 = 24
O = 4 * 2 + 4 = 12
Looking closely at the way I have taken the total number of elements on the reactants and products side, you can solve the rest.
All the Best!
Answer:
5.38 m/s
Explanation:
Given (in the x direction):
Δx = 2.45 m
v₀ = v cos 42.5°
a = 0 m/s²
Δx = v₀ t + ½ at²
(2.45 m) = (v cos 42.5°) t + ½ (0 m/s²) t²
2.45 = (v cos 42.5°) t
t = 3.32 / v
Given (in the y direction):
Δy = 0.373 m
v₀ = v sin 42.5°
a = -9.8 m/s²
Δx = v₀ t + ½ at²
(0.373 m) = (v sin 42.5°) t + ½ (-9.81 m/s²) t²
0.373 = (v sin 42.5°) t − 4.905 t²
0.373 = (v sin 42.5°) (3.32 / v) − 4.905 (3.32 / v)²
0.373 = 2.25 − 54.2 / v²
v = 5.38
Graph:
desmos.com/calculator/5n30oxqmuu
