The lithosphere is Earth's<span> outermost layer, composed of </span>rocks<span> in the </span>crust<span> and upper mantle that behave as brittle solids. The rigid lithosphere sits on top of the asthenosphere, a layer of the mantle in which </span>rocks<span> are hot and deformable.</span>
<h3>
A computer simulation is an example of a <u>Model</u>.</h3>
A model can be a virtual OR physical representation of an object, movement, etc. A model is used to have a visual view of an item.
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In an alpha decay, the original nucleus decays into a daughter nucleus + an alpha particle, which consists of 2 protons and 2 neutrons.
The atomic number (number of protons) of the Francium is 87, and since in the decay it loses 2 protons, the atomic number of the daughter nucleus should be 85. Looking at the periodic table ,this corresponds to Astatine, and therefore this is the element produced in the alpha decay of Francium-221.
Moreover, the mass number (number of protons+neutrons) decreases by 4 (because an alpha particle carries 2 protons+2 neutrons), therefore in the reaction the mass number will decrease from 221 to 217.
So, the daughter element is astatine-217.
A compound machine is a machine composed of two or more simple machines. Common examples are bicycles, can openers and wheelbarrows. Simple machines change the magnitude or direction of a force without any motor. <span>Simple machines are generally easy to understand and work on simple principles. They include such things as levers, which grant mechanical advantage, and wedges, which redirect relative motion, and wheels of various sorts, from pulleys to bicycle gears. </span>
Answer:
b) the second at the doorknob.
Explanation:
The torque applied by a force is given by the equation
where
is the torque
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
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 
, so the torque applied by this first force is
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
Therefore, we see that the 2nd force exerts a greater torque.
