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xxTIMURxx [149]
3 years ago
9

Assuming weightless pulleys and 100% efficiency, what is the minimum input force required to lift a 120 N weight using a single

fixed pulley?
Physics
1 answer:
AnnZ [28]3 years ago
8 0

If a system will consist only one pulley then in that case we will have

F_{input} = F_{output}

here we will say

F_{output} = 120 N

As we know that we need to pull 120 N weight so here we need to apply same force on the other side

<em>So here we can say that on the other side of pulley the applied force must be 120 N </em>

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The estimate of how many stars are in a galaxy is 100 thousand million
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the video identifies the force pair produced when an apple falls through the air. which force belongs in a free-body diagram of
trapecia [35]

The free-body diagram of an apple falling through the air has weight of the apple pointing downwards and the air-resistance on the apple acting upwards.

When an object falls from up to the ground, the object falls under in the influence of acceleration due to gravity.

The vertical component of the force on the apple as it falls trough the air is given as;

∑Fy = 0

Fₙ - W = 0

Fₙ = W

where;

  • <em>Fₙ is the frictional force on the apple acting upwards</em>
  • <em>W is the weight of the apple acting downwards</em>

The free-body diagram of the apple is represented as follows;

                                         ↑ Fₙ

                                         Ο

                                         ↓ W

Thus, the free-body diagram of an apple falling through the air has weight of the apple pointing downwards and the air-resistance on the apple acting upwards.

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6 0
3 years ago
A tourist stands at the top of the Grand Canyon, holding a rock, overlooking the valley below. Find the final velocity and displ
Anestetic [448]

Answer:

a. -39.2 m/s; -78.4 m

b. -31.2 m/s; -46.4 m

c. -47.2 m/s; -110.4 m

Explanation:

<h2>Part (a)</h2>

We are given/can infer these variables:

  • t = 4.0 s
  • a = -9.8 m/s²
  • v_0 = 0 m/s

We want to find the displacement and the final velocity of the rock.

  • Δx = ?
  • v = ?

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = 0 + (-9.8)(4.0)
  • v = -9.8 * 4.0
  • v = -39.2 m/s

The final velocity of the rock is -39.2 m/s.

Now we can use this equation to find the displacement of the rock:

  • Δx = v_0 t + 1/2at²

Plug in the known variables into this equation.

  • Δx = 0 * 4.0 + 1/2(-9.8)(4.0)²
  • Δx = 1/2(-9.8)(4.0)²
  • Δx = -4.9 * 16
  • Δx = -78.4 m

The displacement of the rock is -78.4 m.

<h2>Part (b)</h2>

We are given/can infer these variables:

  • v_0 = 8.0 m/s
  • a = -9.8 m/s²
  • t = 4.0 s

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = 8.0 + (-9.8)(4.0)
  • v = 8.0 + -39.2
  • v = -31.2 m/s

The final velocity of the rock is -31.2 m/s.

We can use this equation to find the displacement:

  • Δx = v_0 t + 1/2at²

Plug in known variables:

  • Δx = 8.0(4.0) + 1/2(-9.8)(4.0)²
  • Δx = 32 - 4.9(16)
  • Δx = -46.4 m

The displacement of the rock is -46.4 m.

<h2>Part (c)</h2>

We are given/can infer these variables:

  • v_0 = -8.0 m/s
  • a = -9.8 m/s²
  • t = 4.0 s

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = -8.0 + (-9.8)(4.0)
  • v = -8.0 - 39.2
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The final velocity of the rock is -47.2 m/s.

We can use this equation to find the displacement:

  • Δx = v_0 t + 1/2at²

Plug in known variables:

  • Δx = -8.0(4.0) + 1/2(-9.8)(4.0)²
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The displacement of the rock is -110.4 m.

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Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i
garik1379 [7]

Answer

given,

wavelength (λ) = 650 nm

angle = 5°

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d= \dfrac{n \lambda}{sin \theta}

d= \dfrac{1 \times 650 \times 10^{-9}\ m}{sin5^0}

d = 7.46 x 10⁻⁴ cm

number of slits per centimeter

  = \dfrac{1}{d}\\\Rightarrow \dfrac{1}{7.46\times 10^{-4}}\\\Rightarrow 1340 split per centimeter.

b) wavelength of two rays  650 nm and 420 nm

 d = \dfrac{1}{5000}

     d =  2 x 10⁻6 m

    we now,

sin \theta = \dfrac{n \lambda}{d}

for 650 nm

sin \theta = \dfrac{2\times 650\times 10^{-9}}{2\times 10^{-6}}

\theta =sin^{-1}(0.65)

θ = 40.54°

for 450 nm

sin \theta = \dfrac{2\times 450\times 10^{-9}}{2\times 10^{-6}}

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θ = 24.83°

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|θ_{650} -θ_{420}| =40.54°-24.83°

|θ_{650} -θ_{420}| =19.71°

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