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yulyashka [42]
3 years ago
6

Does anyone know how to solve this??

Physics
1 answer:
Scrat [10]3 years ago
7 0

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wtjfavyw

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Parallel wave fronts are incident on an opening in a barrier. Which diagram shows the configuration of wave fronts and barrier o
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Explanation :

It is given that, parallel wave fronts are incident on an opening in a barrier. Diffraction of the waves depends on the size of opening and the wavelength of light used as per the following relation :

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Continental deflections, the Coriolis effect and global winds all affect surface ocean currents.
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If the tension in the rope is 160 n, how much work does the rope do on the skier during a forward displacement of 270 m?
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If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.

Given,

Tension force in the rope is (T) = 160 N

Displacement of the skier (S) = 270 m

The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.

Therefore, work done by the rope on  the skier is,

   W=T.S

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A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find t
scoray [572]

Answer:

a)    F₁ = 267.3 N,   N₁ = 1300 N,  b)    μ = 0.324

Explanation:

For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces

           

let's use subscript 1 for the ladder and 2 for the firefighter

            ∑ τ = 0

          -W₁ x₁ - W₂ x₂ + N₁ y = 0

           N₁ = \frac{W_1 x_1 + W_2 x_2}{y}          (1)

the center of mass of the ladder is at its geometric center,

d = L / 2 = 15/2 = 7.5 m

         cos 60 = x₁ / d₁

         x₁ = d₁ cos 60

         x₁ = 7.5 cos 60

         x₁ = 3.75 m

for the firefighter d₂ = 4 m

         cos 60 = x₂ / d₂

         x₂ = d₂ cos 60

          x₂ = 4 cos 60 = 2 m

for the fulcrum d₃ = 15 m

         sin 60 = y / d₃

         y = d₃ sin 60

         y = 15 sin 60

         y = 13 m

we look for the Normal by substituting in equation 1

         N₂ = \frac{500 \ 3.75 \ + 800 \ 2}{13}

         N₂ = 267.3 N

now let's use the translational equilibrium relations

 X axis

           F₁ - N₂ = 0

           F₁ = N₂

           F₁ = 267.3 N

Axis y

          N₁ - W₁ -W₂ = 0

          N₁ = W₁ + W₂

          N₁ = 500 + 800

          N₁ = 1300 N

b) for this case change the firefighter's distance d₂ = 9 m

          x₂ = 9 cos 60

          x₂ = 4.5 m

we substitute in 1

          N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}  

          N₂ = 421.15 N

of the translational equilibrium equation on the x-axis

          fr = F₁ = N₂

          fr = 421.15 N

friction force has the expression

          fr = μ N

in this case the reaction of the Earth to the support of the ladder is N1 = 1300N

          μ = fr / N₁

          μ = 421.15 / 1300

          μ = 0.324

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Answer:

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