- D All linear dimensions within the plate—including the radius and
circumference of the hole—will increase by the same amount during thermal
expansion. (To see that the hole does indeed get bigger, imagine that it was
filled with a flat circular plug of metal. This plug would get bigger as the
entire plate expanded, so if the plug were removed, it would leave behind a
bigger hole.)
- B The net force acting on the block is 80 N − 60 N = 20 N. Dividing the net
force by the object’s mass gives the acceleration (Newton’s second law), so
we find that a = Fnet/m = (20 N)/(40 kg) = 0.5 m/s^2.
- C Since momentum, p, is equal to mv, we just multiply the two entries
(m and v) in each row of the table and see which one is the greatest. This
occurs in Trial 3, where p = (2 kg)(2 m/s) = 4 kg×m/s. So, the answer must be
either (C) or (D). To decide which, we only need to find the kinetic energy of
the object in Trial 2 and Trial 3 and choose the one that’s greater. Since mv^2
= (1 kg)(3 m/s)^2 = J in Trial 2, but only (2 kg)(2 m/s)^2 = 4 J in Trial 3,
we see that the object’s kinetic energy is greater in Trial 2, so the answer is
(C).
- C Choices (A) and (E) were known before Rutherford conducted this series
of experiments, and (B) and (D) concerning the electrons were proposed
later by Bohr. What Rutherford discovered with these experiments was that
an atom’s positive charge was not uniformly distributed throughout the entire
atom but was instead concentrated into a very small volume at the atom’s
center (the nucleus).
- B Use the ideal gas law.
- B We use the equation for the power dissipated by a resistor, P = IV.
