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Answer:
3.982 kg
Explanation:
The latent heat of vaporization = 540 cal/g
= 5.4 ×10⁵ cal/kg
L = 5.4 ×10⁵ × 4.19
= 2.26 × 10⁶ J/kg
Q = 12 × 750,000
= 9, 000, 000
= 9 × 10⁶ J
the maximum number of kg of water (at 100 degrees Celsius) that could be boiled into steam (at 100 degrees Celsius) is:
=
= 3.982 kg
1) In the reference frame of one electron: 0.38c
To find the relative velocity of one electron with respect to the other, we must use the following formula:
where
u is the velocity of one electron
v is the velocity of the second electron
c is the speed of light
In this problem:
u = 0.2c
v = -0.2c (since the second electron is moving towards the first one, so in the opposite direction)
Substituting, we find:
2) In the reference frame of the laboratory: -0.2c and +0.2c
In this case, there is no calculation to be done. In fact, we are already given the speed of the two electrons; we are also told that they travel in opposite direction, so their velocities are
+0.2c
-0.2c