Answer:
0.89 g/cm^3 = 890 kg/m^3
Explanation:
Cross sectional area of U-tube ( A ) = 1.00 cm^2
volume of oil ( V ) = 5.00 cm^3
change between top surface = 0.550 cm
height of oil = 5 cm ( volume / area )
height of water = 5 - 0.550 = 4.45 cm
pressure at the oil-water junction = Pressure on the second side of the U-tube at same level
Po * g * Hoil = Pw * g * Hwater
Po * 5 = 1 * 4.45
∴ Density of oil ( Po ) = 4.45 / 5 g/cm^3 = 0.89 g/cm^3
Answer:
T = 1.766(M-m) Nm where M and m are the 2 masses of the objects
Explanation:
Let m and M be the masses of the 2 objects and M > m so the system would produce torque and rotational motion on the pulley. Force of gravity that exert on each of the mass are mg and Mg. Since Mg > mg, the net force on the system is Mg - mg or g(M - m) toward the heavier mass.
Ignore friction and string mass, and let g = 9.81 m/s2, the net torque on the pulley is the product of net force and arm distance to the pivot point, which is pulley radius r = 0.18 m
T = Fr = g(M - m)0.18 = 0.18*9.81(M - m) = 1.766(M-m) Nm
I believe the answer is D. <span>The hypothesis is revised and another experiment is conducted.</span>
The pressure at a certain depth underwater is:
P = ρgh
P = pressure, ρ = sea water density, g = gravitational acceleration near Earth, h = depth
The pressure exerted on the submarine window is:
P = F/A
P = pressure, F = force, A = area
The area of the circular submarine window is:
A = π(d/2)²
A = area, d = diameter
Set the expressions for the pressure equal to each other:
F/A = ρgh
Substitute A:
F/(π(d/2)²) = ρgh
Isolate h:
h = F/(ρgπ(d/2)²)
Given values:
F = 1.1×10⁶N
ρ = 1030kg/m³ (pulled from a Google search)
g = 9.81m/s²
d = 30×10⁻²m
Plug in and solve for h:
h = 1.1×10⁶/(1030(9.81)π(30×10⁻²/2)²)
h = 1540m
Question:
The water molecules now in your body were once part of a molecular cloud. Only about onemillionth of the mass of a molecular cloud is in the form of water molecules, and the mass density of such a cloud is roughly 2.0×10−21 g/cm^3.
Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.
Answer:
The volume of cloud that has the same density as the amount of water in our body is 1.4×10²⁵ cm³
Explanation:
Here, we have mass density of cloud = 2.0×10⁻²¹ g/cm^3
Density = Mass/Volume
Volume = Mass/Density = If the mass is 40 kg and the body is made up of 70% by mass of water, we have
28 kg water = 28000 g
Therefore the Volume = 28 kg/ 2.0×10⁻²¹ g/cm^3 = 1.4×10¹⁹ m³ = 1.4×10²⁵ cm³.
Therefore, the volume of cloud that has the same density as the amount of water in our body = 1.4×10²⁵ cm³.
