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bearhunter [10]
3 years ago
14

How are contact forces different from non-contact forces?

Physics
2 answers:
madam [21]3 years ago
8 0
A non-contact force is a force which acts on an object without coming physically in contact with it. In contrast a contact force is a force which acts on an object coming physically in contact with it. ...
Sedaia [141]3 years ago
8 0
Contact force needs to be in contact to make “contact force”. Its the opposite for non-contact force, it has to have NO contact to make “non-contact force” happen, i hope this helps
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Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),
salantis [7]

Answer:

The  drag coefficient is  D_z  =  1.30512  

Explanation:

From the question we are told that

     The density of air is  \rho_a  = 1.21 \ kg/m^3

     The diameter of bottom part is  d = 0.15 \ m

The  power trend-line  equation is mathematically represented as

      F_{\alpha }  = 0.9226 * v^{0.5737}

let assume that the velocity is  20 m/s

Then

      F_{\alpha }  = 0.9226 * 20^{0.5737}

       F_{\alpha }  = 5.1453 \ N

The drag coefficient is mathematically represented as

      D_z  =  \frac{2 F_{\alpha } }{A \rho v^2 }

Where  

     F_{\alpha } is the drag force

      \rho is the density of the fluid

       v is the flow velocity

       A is the area which mathematically evaluated as

       A = \pi r^2 =  \pi  \frac{d^2}{4}

substituting values

     A =  3.142 *    \frac{(0.15)^2}{4}

     A = 0.0176 \  m^2

Then

   D_z  =  \frac{2 * 5.1453 }{0.0176 * 1.12 *  20^2 }

   D_z  =  1.30512  

3 0
2 years ago
HELP ASAP PLEASE!!!
Montano1993 [528]

Answer:

The answer is A, B, C, D

Explanation:

This is because gravity is the weakest force of the four fundamental forces, so it automatically cancels letter E

7 0
2 years ago
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Figure 10.20 in your textbook shows an energy diagram for a system with total energy E1. Suppose the system's total energy is E2
wolverine [178]

The particles can undergo small oscillations around x₂.

The given parameters;

  • <em>initial energy of the particles = E₁</em>
  • <em>final energy of the particles, E₂ = 0.33E₁</em>

The movement of the particles depends on the kinetic energy of the particles.

When kinetic energy of the particles is 100%, the particles can oscillate from x₁ to x₅.

However, when the total energy of this particles is reduced to one-third (¹/₃) or 33% of the initial energy of the particle, the oscillation of the particles will be reduced.

  • The maximum position the particle can oscillate is x₅
  • The half position the particles can oscillate is x₃

Since 33% is less than the half of the energy of the particle, the particle will oscillate between x₁ and x₂.

Thus, we can conclude that the particles can undergo small oscillations around x₂.

Learn more here:brainly.com/question/23910777

3 0
2 years ago
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Minchanka [31]

Answer:

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Explanation:

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4 0
3 years ago
Please help although I think I have the awnser but correct me if I’m wrong
iren2701 [21]

I believe you got it correct already

7 0
2 years ago
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