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erik [133]
2 years ago
14

A hill that has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction. At

what angle is such a hill inclined above the horizontal
Physics
1 answer:
Andreas93 [3]2 years ago
6 0

The angle of incline of the hill above the horizontal is 8.81°.

Since the hill has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction.

<h3>Tangent of the angle of the incline of the hill,</h3>

The tangent of the angle of the incline of the hill, Ф is

tanФ = vertical rise/horizontal distance = grade of hill

Now, the vertical rise = 15.5 m and the horizontal distance = 100.0 m

So, substituting the values of the variables into the equation, we have

tanФ = vertical rise/horizontal run

tanФ = 15.5 m/100.0 m

tanФ = 0.155

<h3>Angle of incline of the hill</h3>

Taking inverse tan of both sides, we have

Ф = tan⁻¹(0.155)

Ф = 8.81°

So, the angle of incline of the hill above the horizontal is 8.81°.

Learn more about angle of incline of a hill here:

brainly.com/question/10056962

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The speed of the rock at 20 m is 34.3 m/s

Explanation:

We can solve this problem by using the law of conservation of energy: the mechanical energy of the rock, sum of its potential energy + its kinetic energy) must be conserved in absence of air resistance. So we can write:

U_i +K_i = U_f + K_f

where :

U_i is the initial potential energy

K_i is the initial kinetic energy

U_f is the final potential energy

K_f is the final kinetic energy

The equation can also be rewritten as  follows:

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 100 kg is the mass of the rock

g=9.8 m/s^2 is the acceleration of gravity

h_i = 80 is the initial height

u = 0 is the initial speed  (the rock starts at rest)

h_f = 20 m is the final height of the rock

v is the final speed when h = 20 m

And solving for v, we find:

v=\sqrt{2g(h_i-h_f)}=\sqrt{2(9.8)(80-20)}=34.3 m/s

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