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:
Volume will be 15 mL. Solution:- If we look at the given information then it is Boyle's law as the temperature is constant and the volume changes inversely as the pressure changes. So, the volume of the air bubble at the surface will be 15 mL.
Answer:
31677.2 lb
Explanation:
mass of hammer (m) = 3.7 lb
initial velocity (u) = 5.8 ft/s
final velocity (v) = 0
time (t) = 0.00068 s
acceleration due to gravity (g) 32 ft/s^{2}
force = m x ( a + g )
where
- m is the mass = 3.7 lb
- g is the acceleration due to gravity = 32 ft/s^{2}
- a is the acceleration of the hammer
from v = u + at
a = (v-u)/ t
a = (0-5.8)/0.00068 = -8529.4 ( the negative sign showa the its decelerating)
we can substitute all required values into force= m x (a+g)
force = 3.7 x (8529.4 + 32) = 31677.2 lb
After traveling for 6.0 seconds, a runner reaches 10m/s. What is the runner's acceleration? Answer: 1.67 m/s2