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quester [9]
2 years ago
10

You hang a light in front of your house using an

Physics
1 answer:
Kaylis [27]2 years ago
6 0

The magnitudes of the forces that the ropes must exert on the knot connecting are :

  • F₁ = 118 N
  • F₂ = 89.21 N
  • F₃ = 57.28 N

<u>Given data :</u>

Mass ( M ) = 12 kg

∅₂ = 63°

∅₃ = 45°

<h3>Determine the magnitudes of the forces exerted by the ropes on the connecting knot</h3><h3 />

a) Force exerted by the first rope = weight of rope

∴  F₁ = mg

     = 12 * 9.81 ≈  118 kg

<u>b) Force exerted by the second rope </u>

applying equilibrium condition of force in the vertical direction

F₂ sin∅₂ + F₃ sin∅₃ - mg = 0  ---- ( 1 )

where: F₃ = ( F₂ cos∅₂ / cos∅₃ ) --- ( 2 )  applying equilibrium condition of force in the horizontal direction

Back to equation ( 1 )

F₂ =  [ ( mg / cos∅₂ ) / tan∅₂ + tan∅₃ ]

   = [ ( 118 / cos 63° ) / ( tan 63° + tan 45° ) ]

   = 89.21 N

<u />

<u>C ) </u><u>Force </u><u>exerted by the</u><u> third rope </u>

Applying equation ( 2 )

F₃ = ( F₂ cos∅₂ / cos∅₃ )

    = ( 89.21 * cos 63 / cos 45 )

    = 57.28 N

Hence we can conclude that The magnitudes of the forces that the ropes must exert on the knot connecting are :

F₁ = 118 N, F₂ = 89.21 N, F₃ = 57.28 N

Learn more about  static equilibrium : brainly.com/question/2952156

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Anna007 [38]

Explanation:

its either a or d however i say the best choice is d

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3 years ago
A ray of light crosses a boundary between two transparent materials. The medium the ray enters has a larger optical density. Whi
Tpy6a [65]

Answer:

The wavelength of the light decreases as it enters into the medium with the greater optical density.

The frequency of the light remains constant as it transitions between materials.

Explanation:

- When a ray of light crosses a boundary between two different materials, it undergoes refraction: the ray changes direction, and it also changes speed, according to the relationship:

v=\frac{c}{n}

where v is the speed of the ray of light in the material, c is the speed of light in a vacuum, n is the index of refraction of the material, which is larger for a medium with larger optical density. So, from the equation, we see that the larger the optical density, the smaller the speed of the wave.

- The frequency of the light does not depend on the properties of the medium, so it remains unchanged: therefore the statement

The frequency of the light remains constant as it transitions between materials.

is correct.

- Moreover, the wavelength of the ray of light is related to the speed and the frequency by the equation

\lambda=\frac{v}{f}

where v is the speed and f the frequency. Since we have seen that v decreases and f remains constant, this means that the wavelength decreases as well, so the statement

The wavelength of the light decreases as it enters into the medium with the greater optical density.

is also correct.

5 0
3 years ago
A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re
valentinak56 [21]

Answer:

a) 17.33 V/m

b) 6308 m/s

Explanation:

We start by using equation of motion

s = ut + 1/2at², where

s = 1.2 cm = 0.012 m

u = 0 m/s

t = 3.8*10^-6 s, so that

0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²

0.012 = 0.5 * a * 1.444*10^-11

a = 0.012 / 7.22*10^-12

a = 1.66*10^9 m/s²

If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where

E = electric field

m = mass of proton

a = acceleration

q = charge of proton

E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19

E = 2.77*10^-18 / 1.6*10^-19

E = 17.33 V/m

Final speed of the proton can be gotten by using

v = u + at

v = 0 + 1.66*10^9 * 3.8*10^-6

v = 6308 m/s

5 0
3 years ago
Please answer fast in hindi ​
Ipatiy [6.2K]

Answer:

1. Dheere Dheere (slowly slowly)

2. Har (every)

3. Kal (tomorrow)

4. Mat (don't)

5. Andar (inside)

sorry I wasn't able to write in hindi

8 0
3 years ago
A ball is thrown vertically upwards from a point on the ground. It was observed at a height h twice with a time interval Δt. The
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I don’t think we can answer this question with the information given. ANY ball thrown with ANY initial velocity v will be observed at a height h twice and with a time interval Δt.
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