According to the Law of Conservation of Energy, energy is neither created nor destroyed. They are just transferred from one system to another. To obey this law, the energy of the substances inside the container must be equal to the substance added to it. The energy is in the form of heat. There can be two types of heat energy: latent heat and sensible heat. Sensible heat is energy added or removed when a substance changes in temperature. Latent heat is the energy added or removed at a constant temperature during a phase change. Since there is no mention of phase change, we assume the heat involved here is sensible heat. The equation for sensible heat is:
H = mCpΔT
where
m is the mass of the substance
Cp is the specific heat of a certain type of material or substance
ΔT is the change in temperature.
So the law of conservation of heat tells that:
Sensible heat of Z + Sensible heat of container = Sensible heat of X
Since we have no idea what these substances are, there is no way of knowing the Cp. We can't proceed with the calculations. So, we can only assume that in the duration of 15 minutes, the whole system achieves equilibrium. Therefore, the equilibrium temperature of the system is equal to 32°C. The answer is C.
well it would be A because 55 degrees is going strait well 75 is going literally straight up
Answer:
Amplitude is decreased by a factor of 
if intensity is decreased by a factor of 3.
Explanation:
Intensity of a sound wave is directly proportional to the square of its amplitude.
Therefore, if intensity is 
and amplitude is 
, then
, where, 
is constant of proportionality.
Now, if intensity of sound wave is decreased by a factor of 3. So,
New intensity is, 
Plug in 
for . This gives,
Therefore, amplitude is decreased by a factor of .
Answer:
B. 1700 Hz, 5100 Hz
Explanation:
Parameters given:
Length of ear canal = 5.2cm = 0.052 m
Speed of sound in warm air = 350 m/s
The ear canal is analogous to a tube that has one open end and one closed end. The frequency of standing wave modes in such a tube is given as:
f(m) = m * (v/4L)
Where m is an odd integer;
v = velocity
L = length of the tube
Hence, the two lowest frequencies at which a dog will have increased sensitivity are f(1) and f(3).
f(1) = 1 * [350/(4*0.052)]
f(1) = 1682.69 Hz
Approximately, f(1) = 1700 Hz
f(3) = 3 * [350/(4*0.052)]
f(3) = 5048 Hz
Approximately, f(3) = 5100 Hz
