Answer : The de-Broglie wavelength of this electron, 
Explanation :
The formula used for kinetic energy is,
..........(1)
According to de-Broglie, the expression for wavelength is,
or,
...........(2)
Now put the equation (2) in equation (1), we get:
...........(3)
where,
= wavelength = ?
h = Planck's constant = 
m = mass of electron = 
K.E = kinetic energy = 
Now put all the given values in the above formula (3), we get:
conversion used : 
Therefore, the de-Broglie wavelength of this electron,
I believe the correct response would be true, thermal energy or heat that is produced by friction usually cannot be used to do work.
Potassium is the 19th element so it is B
Answer:
a) F1 = 1999.8 N
, F2 = 4545 N
, F3 = 2778 N
, c) the cans do not collapse because the pressure is applied on both sides
Explanation:
Let's use the pressure equation
P = F / A
Suppose we have atmospheric pressure 1.01 10⁵ Pa
Let's calculate the area of the can that is a parallelepiped
Length L = 25 cm
width a = 18 cm
high h = 11 cm
Side area A = h a
A = 11 18
A1 = 198 10⁻⁴ m²
Lid area
A2 = L a
A2 = 25 18
A2 = 450 10⁻⁴ m²
Other side area
A3 = L h
A3 = 25 11
A3 = 275 10⁻⁴ m²
Now let's calculate the force on these sides
Side 1
F1 = P * A1
F1 = 1.01 10⁵ 198 10⁻⁴
F1 = 1999.8 N
Side 2
F2 = P A2
F2 = 1.01 10⁵ 450 10⁻⁴
F2 = 4545 N
Side 3
F3 P A3
F3 = 1.01 10⁵ 275 10⁻⁴
F3 = 2778 N
We see that the force is greater on side 2 which is where the can should collapse
b) To compare the previous forces we must use the concept of density, in general the cans are made of aluminum that has a density of 2700 kg / m3
d = m / V
m = d * V
V = L a h
V = 0.25 0.18 0.11
V = 0.00495 m3
m = 2700 0.00495
m = 13.4 kg
This is the maximum weight, because much of the volume we calculate is air that has a much lower density
W = 13.4 * 9.8
W = 131.3 N
Let's make the comparison by saying the two magnitudes
Side 1
F1 / W = 1999.8 / 131.3
F1 / W = 15.2
Side 2
F2 / W = 4545 / 131.3
F2 / W = 34.6
Side 3
F3 / W = 2778 / 131.3
F3 / W = 21.2
c) the cans do not collapse because the pressure is applied on both sides: outside and inside, so the net force is zero on each side.
The isotopes of hydrogen used in nuclear reactions? If you mean hydrogen bombs, then the answer is deuterium and tritium, or hydrogen-2 and hydrogen-3.
