You place a box weighing 268.3 N on an inclined plane that makes a 42◦ angle with the horizontal.
1 answer:
Answer:
Wy = - 268.3*cos(42) = - 199.38 [N]
Wx = 268.3*sin(42) = 179.52 [N]
Explanation:
To solve this problem we must make a free body diagram, in the attached image we can see a free body diagram.
Taking the inclined X & y axes with the same angle of 42°, we can see that the weight can be decomposed on both axes.
Since the angle is adjacent to the y-axis, we can use the cosine function
Wy = - 268.3*cos(42) = - 199.38 [N]
Wx = 268.3*sin(42) = 179.52 [N]
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Answer:
<em>0.85c </em>
Explanation:
Rest mass of Kaon = 494 MeV/c²
Rest mass of proton = 938 MeV/c²
The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²
for the kaon, rest energy = 494c² MeV
for the proton, rest energy = 938c² MeV
Recall that the rest energy, and the total energy are related by..
= γ
which can be written in this case as
= γ
...... equ 1
where = total energy of the kaon, and
= rest energy of the kaon
γ = relativistic factor =
where
But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...
=
......equ 2
where is the total energy of the kaon, and
is the rest energy of the proton.
From =
= 938c²
equ 1 becomes
938c² = γ494c²
γ = 938c²/494c² = 1.89
γ = = 1.89
1.89 = 1
squaring both sides, we get
3.57( 1 - ) = 1
3.57 - 3.57 = 1
2.57 = 3.57
= 2.57/3.57 = 0.72
= 0.85
but,
v/c = 0.85
v = <em>0.85c </em>