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Furkat [3]
3 years ago
13

Se coloca una tuerca con una llave, como muestra la figura, si el brazo r= 30 cm y el torque de apriete recomendado para la fuer

za es de 30 Nxm, Cuàl debe ser el valor de la fuerza F aplicada?
Physics
1 answer:
irina [24]3 years ago
6 0

Answer:

F = 100 N

Explanation:

The torque is given by the expression

      τ = F x r

where bold letters indicate vectors, the magnitude of this expression is

     τ = F r sin θ

In general, when tightening a nut, the force is applied perpendicular to the arm, therefore  θ = 90 and sin 90 = 1

      τ = F r

      F = τ / r

calculate

      F = 30 / 0.30

      F = 100 N

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Ossicles

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3 years ago
For the PE formula, why is the height required for calculations? Why do we need to know the height in order to determine PE? *
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Answer:

Answer in Explanation

Explanation:

Whenever we talk about the gravitational potential energy, it means the energy stored in a body due to its position in the gravitational field. Now, we know that in the gravitational field the work is only done when the body moves vertically. If the body moves horizontally on the same surface in the Earth's Gravitational Field, then the work done on the body is considered to be zero. Hence, the work done or the energy stored in the object while in the gravitational field is only possible if it moves vertically. This vertical distance is referred to as height. <u>This is the main reason why we require height in the P.E formula and calculations.</u>

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3 years ago
a projectile is lunched with an initial speed of 60.0mm/s at an angle of 30.0° above the horizontal.The projectile lands on a hi
alexandr402 [8]

Answer:

52 mm/s (approximately)

Explanation:

Given:

Initial speed of the projectile is, u=60.0\ mm/s

Angle of projection is, \theta=30.0\°

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In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.

So, the velocity in the horizontal direction always remains the same.

The horizontal component of initial velocity is given as:

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Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.

Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.

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