Answer:
The angle for the forward Mach line is 19.47°
The angle for the rearward Mach line is 5.21°
Explanation:
From table A-1 (Modern Compressible Flow: with historical perspective):
(M₁ = 3)
If Po₁ = Po₂
Table A-1:
Table A-5:
v₁ = 49.76°
μ₁ = 19.47°
v₂ = 60.55°
μ₂ = 16°
θ = 60.55 - 49.76 = 10.79°
The angle for the forward Mach line is:
μ₁ = 19.47°
The angle for the rearward Mach line is:
θr = μ₂ - θ = 16 - 10.79 = 5.21°
Answer:
C) 6 m/s
Explanation:
Given that
m₁=5000 kg
The initial velocity of 5000 kg car =u₁
m₂=10,000 kg
The initial velocity of 10000 kg car =u₂ = 0 m/s
After collision the final speed of the both car,v = 2 m/s
There is no any external force on the system that is why linear momentum will be conserved.
Linear momentum P = m v
m₁u₁ + m₂u₂ = (m₂ + m₁) v
5000 x u₁ + 10000 x 0 = (5000 + 10000) x 2
5000 x u₁ = 15000 x 2
5 x u₁ = 15 x 2
u₁ = 6 m/s
Therefore the answer is C.
C) 6 m/s
Answer:
accuracy refers to the deviation of a measurement from a standard or true value of the quantity being measured
Answer:
answer is option 4
Explanation:
you have to use option 4 because u need to find out initial velocity (Vi)
Answer:
the time interval that an earth observer measures is 4 seconds
Explanation:
Given the data in the question;
speed of the spacecraft as it moves past the is 0.6 times the speed of light
we know that speed of light c = 3 × 10⁸ m/s
so speed of spacecraft v = 0.6 × c = 0.6c
time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds
Now, from time dilation;
t = Δt₀ / √( 1 - ( v² / c² ) )
t = Δt₀ / √( 1 - ( v/c )² )
we substitute
t = 3.2 / √( 1 - ( 0.6c / c )² )
t = 3.2 / √( 1 - ( 0.6 )² )
t = 3.2 / √( 1 - 0.36 )
t = 3.2 / √0.64
t = 3.2 / 0.8
t = 4 seconds
Therefore, the time interval that an earth observer measures is 4 seconds
