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Whitepunk [10]
3 years ago
5

Consider equal volumes (say 1 l of a given substance in the solid, liquid, and gas phases. arrange them in order of decreasing m

ass based on the trend for the average substance. if the mass difference between samples is relatively small (10% or less, rank the items as equivalent.
Physics
2 answers:
lana [24]3 years ago
6 0
The particles in solids are closely packed together, which means that the amount of substance in solids are more compared to the liquid and gases of having the same volume. The arrangement of the phases in decreasing masses will then be,
                                      solid, liquid, and gas
Ad libitum [116K]3 years ago
4 0

Answer:

In the common cases, solids have all the particles close together and held in place, liquids have the particles close together (less than solids) but not held, and gases have the particles more separated than the previous cases.

So, in a defined volume, usually, solids will have more mass, this is because particles are close together and you can put more particles in a given volume, then comes liquid, and the comes to gas.

An exception of this can be water, where the solid-state is less dense than the liquid state (this is why ice floats in liquid water)

Here for a fixed volume, liquid water has more mass than the solid-state, and the solid-state has more mass than the gas state.

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levacccp [35]

Answer:

a) v₂ = 30 m/s

b) m₁ = 12600 kg

c) m₂ = 12600 kg

Explanation:

a)

Using the continuity equation:

A_1v_1 = A_2v_2

where,

A₁ = Area of inlet = π(0.15 m)² = 0.07 m²

A₂ = Area of outlet = π(0.05 m)² = 0.007 m²

v₁ = speed at inlet = 3 m/s

v₂ = speed at outlet = ?

Therefore,

(0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}

<u>v₂ = 30 m/s</u>

<u></u>

b)

m_1 = \rho A_1v_1t

where,

m₁ = mass of water flowing in = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\

<u>m₁ = 12600 kg</u>

<u></u>

c)

m_1 = \rho A_1v_1t

where,

m₂ = mass of water flowing out = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\

<u>m₂ = 12600 kg</u>

7 0
3 years ago
A T-Bar is similar to a_______
dezoksy [38]

Answer:

Donkey

Explanation:

3 0
3 years ago
A positive statement is:________. a. reflects oneâs opinions. b. can be shown to be correct or incorrect. c. a value judgment. d
kenny6666 [7]

Answer:

b

Explanation:

8 0
3 years ago
Which of the choices is not a statement or direct application of the second law of thermodynamics? There are no 100% efficient h
Nat2105 [25]

Answer:

Explanation:

Heat energy naturally transfers from a high temperature substance to a low temperature substance.

It  is not a statement or direct application of the second law of thermodynamics.

The change in internal energy of a system can be found by combining the heat energy added to a system minus the work done by the system.

It  is not a statement or direct application of the second law of thermodynamics. This statement is in accordance with first law of thermodynamics.

All the other two statements are in accordance with second law of thermodynamics.

3 0
3 years ago
A common cylindrical copper wire used in a lab is 841 m long. Find the radius (in mm) of a wire necessary to have 0.5 Ohms of re
slava [35]
The relationship between the resistance R of a wire and its resistivity \rho is given by
R=  \frac{\rho L}{A}
where L is the length of the wire and A is its cross sectional area.

In the problem, we have R=0.5 \Omega, \rho = 1.68 \cdot 10^{-8} \Omega m and L=841 m. So we can solve the find the area A:
A= \frac{\rho L}{R}=2.83 \cdot 10^{-5} m^2

For a cylindrical wire, the cross sectional area is given by
A= \pi r^2
where r is the radius. We know the value of the area A, so now we can find the radius of the wire:
r= \sqrt{ \frac{A}{\pi} }= \sqrt{ \frac{2.83 \cdot 10^{-5}m^2}{\pi} }=0.003 m=3.0 mm
3 0
3 years ago
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