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AfilCa [17]
3 years ago
13

Two equal forces are applied to a door. The first force is applied at the midpoint of the door; the second force is applied at t

he doorknob. Both forces are applied perpendicular to the door. Which force exerts the greater torque?a) the first at the midpoint.
b) the second at the doorknob.
c) both exert equal non-zero torques.
d) both exert zero torques.
e) Additional information is needed.
Physics
1 answer:
Olenka [21]3 years ago
5 0

Answer:

b) the second at the doorknob.

Explanation:

The torque applied by a force is given by the equation

\tau = F_p d

where

\tau is the torque

F_p is the component of the force perpendicular to the direction between the axis of rotation and the point of application of the force

d is the arm (the distance between the axis of rotation and the point of application of the force)

In this problem, we have two equal forces F both applied perpendicular to a door, so

F_p = F

The first force is applied at the midpoint of the door; if we call L the width of the door, then the arm of this force is d_1=\frac{L}{2}, so the torque applied by this first force is

\tau_1 = F\frac{L}{2}

The 2nd force instead is applied at the doorknob, so the arm in this case is L:

d = L

So, the torque exerted by the second force is

\tau_2 = FL

Therefore, we see that the 2nd force exerts a greater torque.

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Answer:

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Explanation:

The isotope decay of an atom follows the equation:

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