The meat in the freezer is frozen.
Everything else in the freezer is frozen too.
Nothing in the refrigerator is frozen.
The freezer is colder than the refrigerator. <span>
Mildred takes a pound of frozen hamburger meat out of the freezer
and puts it into the refrigerator. The meat is colder than anything
else that's in there.
Heat flows from the air in the refrigerator into the frozen hamburger (C)
and warms up the meat. When the temperature of the meat warms up
to the temperature of the air in the refrigerator, the heat stops flowing.</span>
In vacuum and empty space, it's 299,792,458 meters per second
(186,282.4 miles per second).
In any material stuff, it's somewhat less. How much less depends on
what stuff it is ... it's different in each material "medium".
Answer:
2.25 %
Explanation:
65-95-99.7 is a rule to remember the precentages that lies around the mean.
at the range of mean (
) plus or minus one standard deviation (
),
at the range of mean plus or minus two standard deviation,
at the range of mean plus or minus three standard deviation,
So, note that they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) (
)
So we know that the 95.5% is between
and
, hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than 0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.
So
As we know that with respect to oxygen atom taken as reference the product of atomic mass and specific heat of a metal will remain constant.
this product is equal to 0.38
so here we will say that let atomic mass of the metal is M
so with respect to oxygen atom its mass is given as

now we will have

now we will have

so atomic mass of the metal is 7 g/mol
Answer:
a.241.08 m/s b. 196 Hz c. 392 Hz
Explanation:
a. Determine the speed of waves within the wire.
The frequency of oscillation of the wave in the string, f = nv/2L where n = harmonic number, v = speed of wave in string, L = length of string = 1.23 m.
Since f = 588 Hz which is the 6 th harmonic, n = 6. So, making v subject of the formula, we have
v = 2Lf/n
substituting the values of the variables into v. we have
v = 2 × 1.23 m × 588Hz/6
v = 241.08 m/s
b. Determine the frequency at which the wire will vibrate with the first harmonic wave pattern.
The first harmonic is obtained from f when n = 1,
So, f = v/2L = 241.08 m/s ÷ 1.23m = 196 Hz
c. Determine the frequency at which the wire will vibrate with the second harmonic wave pattern.
The second harmonic f' = 2f = 2 × 196 Hz = 392 Hz