The right formula to use for this calculation is the heat capacity formula,
Heat absorbed, Q = MCT, Where
M = Mass of the substance = 71
C = Specific heat capacity for iron = 0.450 J/gc
T = Change in temperature = 29 - 11 = 18
Q = 71 * 0.450 *18 = 575.10
The amount of heat absorbed by the iron skillet is 575 J.
How do you get out of a room with only a mirror and a table?
Look in the mirror you see what you saw, use the saw to cut the table, half and half makes a hole, jump in it.
Thought you would like that.
Answer:
0.67 s
Explanation:
This is a simple harmonic motion (SHM).
The displacement,
, of an SHM is given by

A is the amplitude and
is the angular frequency.
We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or
radian.
From trigonometry,
if A and B are complementary.
At
, 


So

At
, 





The period,
, is related to
by
