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harkovskaia [24]
3 years ago
11

What must happen for salt and water to form a solution?

Physics
1 answer:
Basile [38]3 years ago
6 0
The answer would be D
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In which case does viscosity play a dominant role? Case A: a typical bacterium (size ~ 1 mm1 mm and velocity ~ 20 mm/s20 mm/s) i
My name is Ann [436]

Answer:

Case A

Explanation:

given,

size of bacteria = 1 mm x 1 mm

velocity = 20 mm/s

size of the swimmer = 1.5 m x 1.5 m

velocity of swimmer = 3 m/s

Viscous force

F = \eta A \dfrac{dv}{dx}

for the bacteria

F = \eta \times 10^{-6}\times 20\times 10^{-3}

F =2\times 10^{-8} \eta\ N

for the swimmer

F = \eta \times 1.5^2\times 3

F =6.75 \eta\ N

from the above force calculation

In case B inertial force that represent mass is more than the inertial force in case of bacteria.

Viscous force is dominant in case of bacteria.

So, In Case A viscous force will be dominant.

5 0
3 years ago
Which statement about ammeters and voltmeters is true?
Morgarella [4.7K]
What are the staments
7 0
3 years ago
Water flows through a horizontal nozzle in steady flow at the rate of 10m3/s. The inlet and outlet diameters are d1 = 0.5m and d2
Dvinal [7]

Answer:

P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,

Explanation:

This problem of fluid mechanics let's start with the continuity equation to find the speed of water output

        Q = A v

        v = Q / A

The area of ​​a circle is

       A = π r² = π d² / 4

Let's look at the speeds at each point

       v₁ = Q / A₁ = Q 4 /π d₁²

       v₁ = 10 4 /π 0.5²

       v₁ = 50.93 m / s

       v₂ = Q / A₂

       v₂ = 10 4 /π 0.25²

       v₂ = 203.72 m / s

Now we can use Bernoulli's equation in the colon

       P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear

       P₁ = P2 + ½ rho (v₂² - v₁²)

      P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)

      P₁ = 1.013 10⁵ + 2.205 10⁷

      P₁ = 2.215 10⁷ Pa

la definicion de presion es

      P₁ = F₁/A₁

     F₁ = P₁ A₁

     F₁ = 2.215 10⁷ pi d₁²/4

     F₁ = 2.215 10⁷ pi 0.5²/4

     F₁ = 4.3 106 N

     

6 0
3 years ago
2. Three blocks, A,B and C of mass 2kg. 3kg. 5kg respectively kept side by side with one another are accelerated at 2m/s2 across
gulaghasi [49]

Answer:

Total mass of combination = 2+3+5 = 10kg.

Acceleration produced = 2m/s^2

hence force =( total mass × acceleration)= (2×10)= 20 N.

Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N

applied force on 2 kg block = 20N

Force between 2 kg and 3 kg block = (20-4) = 16N. ans

Net force on 3 kg block = 3 × 2 =6N.

Applied force on 3 kg block due to 2 kg block = 16N.

hence, force between 3 kg and 5 kg block = (16-6) = 10N .

answers:-

(a) 20 N

(b) 16N

(c) 10 N

4 0
2 years ago
A circuit is constructed with six resistors and two batteries as shown. the battery voltages are v1 = 18 v and v2 = 12 v. the po
VladimirAG [237]

Answer:

V4=9.197v

Explanation:

Given:

V1= 18v ,V2= 12v ,r1=r5=58ohms ,r2=r6=124ohms , r3=47ohms ,r4= 125ohms

V4= I4R4 = V2/(R4 + R5)×R4

V4= 12×125 /(125 + 58)

V4=1500/183 =9.197v

5 0
3 years ago
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