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2 years ago

Explain How do the attractions between particles in a fluid determine viscosity?

1 answer:

6 0

Answer: Viscosity is due to particle mass, particle shape, and the strength of the attraction between the particles of a liquid. In general, the stronger the attractive forces between particles, the higher the viscosity. For many liquids, viscosity decreases when heat is applied.

Explanation:

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The amount of space between two points is measure in unit?

Andreas93 [3]

Answer:

meters

Explanation:

I'm not positive if this is correct, your teacher may be looking for a broader answer so possibly just 'distance'. Hope this helps! <3

3 0

2 years ago

Read 2 more answers

For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro

viktelen [127]

Answer:

$V_d$ = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, $A = \frac{\pi d^2}{4}$ = $A = \frac{\pi 0.625^2}{4}$

Now,

The current density, J is given as

$J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}$ = 2444619.925 A/mm²

now, the electron density, $n = \frac{\rho}{M}N_A$

where,

$N_A$ =Avogadro's Number

$n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3$

Now,

the drift velocity, $V_d$

$V_d=\frac{J}{ne}$

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

$V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e}$ = 1.75 × 10⁻⁴ m/s

4 0

2 years ago

Read 2 more answers

The gravitational force between two objects will be greatest in which of the following situations?

Kobotan [32]

Answer:

Explanation:

Gravitational law states that, the force of attraction or repulsion between two masses is directly proportional to the product of the two masses and inversely proportional to the square of their distance apart.

So,

Let the masses be M1 and M2,

F ∝ M1 × M2

Let the distance apart be R

F ∝ 1 / R²

Combining the two equation

F ∝ M1•M2 / R²

G is the constant of proportional and it is called gravitational constant

F = G•M1•M2 / R²

So, to increase the gravitational force, the masses to the object must be increased and the distance apart must be reduced.

So, option c is correct

C. Both objects have large masses and are close together.