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Ugo [173]
3 years ago
7

What is the length of the hypotenuse of a right triangle, whose legs have lengths of 12 meters and 35 meters?

Physics
2 answers:
djyliett [7]3 years ago
8 0

Answer:

Hypotenuse, h = 37 meters

Explanation:

In a right angled triangle, the length of legs are 12 meters and 35 meters. We have to find the length of the hypotenuse of a right angled triangle.

It can be calculated using Pythagoras theorem :

(hypotenuse)^2=(base)^2+(perpendicular)^2

So, (hypotenuse)^2=(12)^2+(35)^2

hypotenuse = 37 meters.

Hence, the length of the hypotenuse of a right angled triangle is 37 meters.                                                    

Makovka662 [10]3 years ago
5 0
The answer is 37 cause the two sides squared and added together equals the hypotenuse squared after that you just find the square root of the answer
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Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet
Valentin [98]

Answer:

v = 8.09   m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

       

starting point. Higher

         Em₀ = U = m gh

final point. To go down the slope

         Em_f = K = ½ m v²

The work of the friction force is

         W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

        N - W_y = 0

        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

         θ = sin⁻¹  0.1

          θ = 5.74º

         sin 5.74 = Wₓ / W

         cos 5.74 = W_y / W

         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

         fr = μ W cos θ

Work is friction force is

         W_fr = - μ W L cos θ  

Let's use the relationship of work with energy

        W + ΔU = ΔK

         -μ mg L cos 5.74 + (mgh - 0) = 0  - ½ m v²

        v² = - 2 μ g L cos 5.74 +2 (gh)

        v² = 2gh - 2 μ gL cos 5.74

let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

        v² = 215.6 -150.16

        v = √65.44

        v = 8.09   m/s

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2 years ago
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