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Marrrta [24]
2 years ago
14

n open rectangular tank is filled to a depth of 2 m with water (density 1000 kg/m3). On top of the water there is a 1 m deep lay

er of gasoline (density 700 kg/m3). The width of the tank is 1 m (the direction perpendicular to the paper). The tank is surrounded by air at atmospheric pressure. Calculate the total force on the right wall of the tank, and specify its direction. The acceleration of gravity g = 9.81 m/s2.
Physics
1 answer:
Marina CMI [18]2 years ago
3 0

Answer:

gasoline zone    P_net = 6860 Pa

Water zone        P_net = 26460 Pa

Force direction is out of tank  

Explanation:

The pressure is defined

         P = F / A

         F = P A

let's write Newton's equation of force

            F_net = F_int - F_ext

            P_net A = (P_int - P_ext) A

             

The P_ext is the atmospheric pressure

         P_ext = P₀

the pressure inside is

gasoline zone

         P_int = P₀ + ρ' g h'

         

water zone

         P_int = P₀ + ρ' g h' + ρ_water h_water

     

we substitute

Zone with gasoline

            P_net = ρ' g h'

            P_net = 700 9.8  1

            P_net = 6860 Pa

Water zone

             P_net = rho’ g h’ + rho_water g h_water

             P_net = 6860 + 1000 9.8  2

             P_net = 26460 Pa

To find the explicit value of the force, divide by a specific area.

Force direction is out of tank

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A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most
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Answer:

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A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

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Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

      I_{total}=I_{body} + 2 I_{arm}

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The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

       M = 7/8 m total

       M = 7/8 64

       M = 56 kg

The mass of the arms is

      m’= 1/8 m total

      m’= 1/8 64

      m’= 8 kg

As it has two arms the mass of each arm is half

     m = ½ m ’

     m = 4 kg

The arms are very thin, we will approximate them as a particle

    I_{arm} = M D²

Let's write the equation

     I_{total} = ½ M R² + 2 (m D²)

Let's calculate

    I_{total} = ½ 56 0.20² + 2 4 0.20²

    I_{total} = 1.12 + 0.32

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b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

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From conservation of energy

K1 +U1 = K2 +U2

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0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

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3 years ago
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Lapatulllka [165]

Answer:

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a=1.283\ m.s^{-2} downwards

Explanation:

Given:

weight of the person, w=688\ N

So, the mass of the person:

m=\frac{w}{g}

m=\frac{688}{9.81}

m=70.132\ kg

  • Now if the apparent weight in the elevator, w_a= 726\ N

<u>Then the difference between the two weights is :</u>

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\Delta w=726-688

\Delta w=38\ N is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

a=\frac{\Delta w}{m}

a=\frac{38}{70.132}

a=0.5418\ m.s^{-2} upwards, because the normal reaction that due to the weight of the body is increased here.

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<u>Then the difference between the two weights is :</u>

\Delta w=w-w_a

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a=\frac{\Delta w}{m}

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7 0
3 years ago
Read 2 more answers
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