M = 10.0 g, the mass of the iron sample
ΔT = 75 - 25.2 = 49.5°C, the decrease in temperature
c = 0.449 J/(g-°C), the specific heat of iron
The heat released is
Q = m*c*ΔT
= (10.0 g)*(0.449 J/(g-°C))*(49.5 C)
= 222.255 J
Answer: 222.3 J (nearest tenth)
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance

Where, d = distance
A = amplitude
Put the value into the formula


Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
(Hint: the time<span> to rise to the </span>peak<span>is one-half the </span>total hang-time<span>.).</span>
Hi there!
The period of an orbit can be found by:

T = Period (? s)
r = radius of orbit (6400000 m)
v = speed of the satellite (8000 m/s)
This is the same as the distance = vt equation. The total distance traveled by the satellite is the circumference of its circular orbit.
Let's plug in what we know and solve.

The liquid that will evaporate the quickest would be water.
<em>~The rest are mixtures, and have more ingrediants in them, therefore the answer should be water. (in which it has no other ingrediants in it but water itself)</em>