Answer:
1.52 seconds
Explanation:
Step 1: identity the given parameters
Initial velocity (u) = 12m/s
Height above ground (h1) = 4m
Final velocity (V) = 0
Step 2: calculate the height travelled by the object from 4m height (h2).
V^2 = U^2 -2gh
0= 12^2-2(9.8*h)
2(9.8*h) = 12^2
19.6*h = 144
h = 144/19.6
h = 7.347 m
Total height above ground (ht) = 4m +7.347m = 11.347m
Step 3: calculate the time reach ground
T = √(2h/g)
T = √(2*11.347/9.8)
T= √(22.694/9.8)
T= √2.316
T= 1.52 seconds
Answer:
Explanation:
because it is a function of physics that will always be true, hope this helps :)
Answer:
c is the answer because it i in a series not a parallel circut.
Explanation:
Answer:
100 cm³
Explanation:
Use ideal gas law:
PV = nRT
where P is absolute pressure, V is volume, n is number of moles, R is ideal gas constant, and T is absolute temperature.
n and R are constant, so:
P₁V₁/T₁ = P₂V₂/T₂
If we say point 1 is at 40m depth and point 2 is at the surface:
P₂ = 1.013×10⁵ Pa
T₂ = 20°C + 273.15 = 293.15 K
P₁ = ρgh + P₂
P₁ = (1000 kg/m³ × 9.8 m/s² × 40 m) + 1.013×10⁵ Pa
P₁ = 4.933×10⁵ Pa
T₁ = 4.0°C + 273.15 = 277.15 K
V₁ = 20 cm³
Plugging in:
(4.933×10⁵ Pa) (20 cm³) / (277.15 K) = (1.013×10⁵ Pa) V₂ / (293.15 K)
V₂ = 103 cm³
Rounding to 1 sig-fig, the bubble's volume at the surface is 100 cm³.
