What is jupiters orbital period
1 answer:
You might be interested in
Answer:
121.3 cm^3
Explanation:
P1 = Po + 70 m water pressure (at a depth)
P2 = Po (at the surface)
T1 = 4°C = 273 + 4 = 277 K
V1 = 14 cm^3
T2 = 23 °C = 273 + 23 = 300 K
Let the volume of bubble at the surface of the lake is V2.
Density of water, d = 1000 kg/m^3
Po = atmospheric pressure = 10^5 N/m^2
P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2
Use the ideal gas equation
By substituting the values, we get
V2 = 121.3 cm^3
Thus, the volume of bubble at the surface of lake is 121.3 cm^3.
Answer:
- path differnce = 2.18*10^-6
- 1538 lines
Explanation:
- The path difference for the waves that produce the pattern of diffraction, is given by the following formula:
(1)
d: separation between slits = 0.50mm = 0.50*10^-3 m
θ: angle of a diffraction = 0.25°
Then, the path difference is:
- The maximum number of bright lines are calculated by using the following formula:
(2)
m: order of the bright
λ: wavelength = 650nm
The maximum bright is calculated for an angle of 90°:
The maxium number of bright lines are twice the previous result, that is, 1538 lines