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lana [24]
2 years ago
11

Ca anyone help me with writing a lab report about the reaction between the pepper and the soap in water

Physics
1 answer:
7nadin3 [17]2 years ago
6 0

Answer:

Buoyancy force and surface tension are the reactions that take places between soap and pepper experiment.

Explanation:

Surface tension:

The surface tension of a liquid is the tendency of liquid surfaces to resist an external force due to the cohesive nature of its molecules.

The pepper and soap experiment helps you to understand buoyancy force and surface tension.

Reaction between the pepper and soap is as following.

  • The pepper flakes float because of buoyancy force. It makes the pepper flakes to move away to the edge of the plate.
  • This happens because the liquid dish soap changes the surface tension of water.
  • And The pepper flakes are so light, it floats on the water surface due to surface tension.
  • when we add soap, it breaks the surface tension of water, but the water resists it. So they pull away from the soap along with the pepper flakes.
  • This pushes the pepper away from your soap covered finger.

This is the reaction that take places between soap and pepper experiment.

Learn more about Pepper and soap experiment here:

<u>brainly.com/question/9614070</u>

<u>#SPJ4</u>

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Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

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and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

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\boxed{\alpha_1 > \alpha_2.}

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Explanation:

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a)

To answer this part, we re-arrange the previous equation as follows:

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this means that if the mass increases, the acceleration decreases, and if the mass decreases, the acceleration increases.

In a graph of acceleration vs mass, the curve representing this relationship would be a hyperbole. The graph is shown as the first graph in the attached picture.

b)

As before, we re-arrange the previous equation as follows:

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Here we notice that if the object has a constant mass, the acceleration is directly proportional to the force applied on the object:

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In a graph of acceleration vs force, the curve representing this relationship would be a straight line, as shown in the second graph in the attached picture.

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