Hey people text me at azieyafreeman2 at g mail dot com
2 answers:
You might be interested in
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>
Answer:
108.37°C
Explanation:
P₁ = Initial pressure = 101 kPa
V₁ = Initial volume = 530 m³
T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K
P₂ = Final pressure = 101 kPa (because it is open to atmosphere)
V₂ = Final volume = 530 m³
P₁V₁ = n₁RT₁
⇒101×530 = n₁RT₁
⇒53530 J = n₁RT₁
P₂V₂ = n₂RT₂
⇒53530 J = n₂RT₂
Dividing the first two equations we get
∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C