The pressure gauge shown below has a spring for which k = 60 n/m, and the area of the piston is 0.50 cm2. its right end is connected to a closed container of gas at a gauge pressure of 30 kpa. how far will the spring be compressed if the region containing the spring is (a) in vacuum, and (b) open to the atmosphere? atmospheric pressure is 101 kpa; *use hooke’s law*

a) -0.1092 m

b) - 0.00283 m

Explanation:

Given:

A_p = 0.5 cm^2 = 0.00005 m^2

P_g = 30 KPa , P_abs = 30KPa + 101 KPa = 131 KPa

k = 60 N /m

Equations used:

F_net = Δ P * A_p             Δ P: Absolute Pressure

F_net = k*x (Hooke's Law)

x = Δ P * A_p  / k

part a (P_ext = 0 Pa)

x = ((0 - 131)*10^3 * 0.00005) / 60 = - 0.1092 m (compression)

part b (P_ext = 101 KPa)

x = ((101 - 131)*10^3 * 0.00005) / 60 = - 0.002583 m (compression)

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.