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algol [13]
2 years ago
8

a 20N mass is supported by two ropes. what is the tension in each rope? how woould i work this problem if i know the two angles

and thats the only info i know.
Physics
1 answer:
TiliK225 [7]2 years ago
5 0

Before we could discuss this in any specific detail, I think we would have to
know the angles.  A generic discussion without actual numbers for the angles
would be just plain too confusing.

The general approach is that the vertical components of both tensions
add up to 20N, and the horizontal components are equal but in opposite
directions.  That's the only way that the mass is hanging motionless.

You have to find the horizontal and vertical components of the tensions
by using the angles and maybe the lengths of the ropes.


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Because of surface tension, it is possible, with care, to support an object heavier than water on the water surface. The maximum
boyakko [2]

Answer:

\pi 1 = \frac{h}{l}

\pi 2 = б / l^2gp

\frac{h}{l} = Ф ( б / l^2gp )

Explanation:

<u>Develop a suitable set of dimensionless parameters for this problem</u>

The set of dimensionless parameters for this problem is :

\pi 1 = \frac{h}{l}

\pi 2 = б / l^2gp

\frac{h}{l} = Ф ( б / l^2gp )

and they are using the pi theorem, MLT systems

attached below is a detailed solution

7 0
3 years ago
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Ivahew [28]
That would be position Y, as the northern hemisphere is tilted away from the sun.
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How much meters is a mile
elena-s [515]

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6 0
3 years ago
Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.
vagabundo [1.1K]

Answer:

R_total = 14.57 Ω ,  V_C = 1.176 V

Explanation:

To solve this circuit we are going to find the equivalent resistance of each branch, let's remember

* Serial resistance  

         R_{eq} = ∑ R_{i}

* For resistance in parallel

        1 / R_{eq} = ∑ 1/R_{i}

We solve the two branches of the wheatstone bridge

Series resistors

Branch B

         R_B = Rb + R4

         R_B = 2 + 18

         R_B = 20 Ω

Branch C

         R_C5 = Rc + R5

         R_C5 = 3 + 12

         R_C5 = 15 Ω

Resistance in parallel R_B and R_C5

         1 / R_BC = 1 / R_B + 1 / R_C5

          1 / R_BC = 1/20 + 1/15 = 0.116666

          R_BC = 8.57 Ω

Now we have a single branch, we solve the series resistance

          R_total = R_A + R_BC

          R_total = 6 + 8.57

          R_total = 14.57 Ω

b) they ask us for the voltage in the resistance R_C

Let's remember that the voltage in a series circuit is the sum of the voltages

           10 = V_a + V_BC

           10 = i R_a + i R_BC = i (R_a + R_BC)

           i = 10 / (R_a + R_BC)

           i = 10 / (14.57)

           i = 0.6863 A

The current in the series circuit is constant

          V_BC = i R_BC

          V_BC = 0.6863 8.57

          V_BC = 5.8819 V

This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.

     Branch C

             V_BC = i R_C5

             i = V_BC / R_C5

             i = 5.8819 / 15

             i = 0.39213 A

In this branch we have two resistors in series, let's remember that the current in a series circuit is constant

             V_C = i R_C

              V_C = 0.39213 3

              V_C = 1.176 V

3 0
3 years ago
The charge on the sphere is monitored as a beam of monochromatic light shines on the sphere. Initially nothing happens. The wave
Solnce55 [7]

Answer:

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Explanation:

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the energy of the photons is not enough to carry out an electronic transition between two states of the material, when we decrease the wavelength (the energy of the photons increases), the point is reached where the energy of the beam is equal to some energy of a transition, by which the electrons are promoted and since we can see a certain charge, as the atoms are neutral, some electrons must be removed from the material, this is represented in the macroscopic case as the work function of the material, consequently a unbalanced load that is what we can measure.

When we increase the lightning intensity, what we do is that we increase the number of photons and if each photon can remove an electron, by removing the electrons the difference between it and the positive charge (fixed in the nuclei) increases.

We can analyze the interaction of the photon and the electron as a particular collision.

The explanation of this effect was made by Einstein in his explained of the photoelectric effect

8 0
2 years ago
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