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.
They will rise to the 2nd layer of the atmosphere where the temperature decreases by a lot and then they will blow up
Answer:
a) 1504.8 J
b) 991.76 J
c) 0J
d) 0J
Explanation:
(a) The work done by the force P on the box is given by the following formula:

P: applied force = 171N
x: distance in which the for P is applied = 8.80m
you replace the values of P and x and obtain:

(b) The work don by the friction force is:

μ = coefficient of kinetic friction = 0.250
M: mass of the box = 46.0kg
g: gravitational constant = 9.8 m/s^2

(c) The Normal force is

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

(d) the same as before:

Answer:
Described below
Explanation:
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Density = mass/volume
= 4300/10000
= 0.43 g/mL.
Therefore density of liquid propane is 0.43 g/mL.
Hope this helps!