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Yuliya22 [10]
3 years ago
13

Suppose you were asked to find the torque about point p due to the normal force n in terms of given quantities. which method of

finding the torque would be the easiest to use?

Physics
1 answer:
igomit [66]3 years ago
7 0

Answer:

T=Lnsin\alpha

Please check the attached

Explanation:

The torque can simply be calculated by multiplying the length of the rod by the perpendicular force n as shown in the attached figure.

Note that sin90=1

T=Lsin\alpha(nsin90)

T=Lsin\alphaxn

T=Lnsin\alpha

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Which of these will have the same units of pressure is multiplied by area? Velocity time force or pressure
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Considering the definition of pressure, force will have the same units of pressure is multiplied by area.

<h3>What is pressure</h3>

Pressure is defined as the force exerted per unit area.

In this way, for the same force, the pressure is inversely proportional to the surface: this means that, for the same force, the smaller the surface, the greater the pressure exerted and the more the body will be deformed.

That is, pressure is defined as the force exerted perpendicularly on a surface and is expressed mathematically as:

Pressure=\frac{Force}{Area}

In the International System of Units, the unit of pressure is the pascal (Pa). One pascal is the pressure exerted by a force of 1 N distributed over an area of ​​1 m².

Then, the definition of force can be expressed as:

Force= Pressure× Area

<h3>Summary</h3>

In summary, force will have the same units of pressure is multiplied by area.

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The trajectory of a rock ejected from the Kilauea volcano, with a velocity of magnitude 6.4 m/s and at angle 2.2 degrees above t
Vika [28.1K]

Answer:

v = 8.03 m / s

Explanation:

This is a missile throwing exercise.

          y = y₀ + v_{oy}  t - ½ g t²

     

indicates that y = -1.2 m and the initial velocity is

         v_{oy} = v₀ sin θ

         v_{oy} = 6.4 sin 2.2

         v_{oy} = 0.2457 m / s

we substitute in the equation

         -1.2 = 0.2457 t - ½ 9.8 t²

          4.9 t² - 0.2457 t - 1.2 = 0

          t² - 0.05014 t -0.2449 = 0

we solve the quadratic equation

          t = [0.05014 ±√ (0.05014² + 4 0.2449)] / 2

          t = [0.05014 ± 0.9910] / 2

          t₁ = 0.5206 s

          t₂ = -0.47 s

time must be a positive magnitude therefore the correct answer is

          t = 0.5206 s

with this time we can calculate the vertical speed when the rock hits the ground

         v_{y} = v_{oy} - gt

         v_{y} = 0.2457 - 9.8 0.5206

         v_{y} = -4.856 m / s

the negative sign indicates that the speed is down

horizontal velocity is constant, due to no acceleration

         vₓ = v₀ₓ = v₀ cos 2,2

         v₀ₓ = 6.4 cos 2.2

         v₀ₓ = 6.395 m / s

therefore let's use Pythagoras' theorem to find the velocity

         v = √ (vₓ² + v_{y}^{2})

         v = √ (6,395² + 4,856²)

         v = 8.03 m / s

the direction can be found with trigonometry

        tan θ = v_{y} / vₓ

        θ = tan⁻¹ (-4,856 / 6,395)

        θ = - 37

the negative sign indicates that it is half clockwise from the x axis

   

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