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
Answer:
The sled slides d=0.155 meters before rest.
Explanation:
m= 60 kg
V= 2 m/s
μ= 0.3
g= 9.8 m/s²
W= m * g
W= 588 N
Fr= μ* W
Fr= 176.4 N
∑F = m * a
a= (W+Fr)/m
a= 12.74m/s²
t= V/a
t= 0.156 s
d= V*t - a*t²/2
d= 0.155 m
From tables, the speed of sound at 0°C is approximately
V₁ = 331 m/s (in air)
V₃ = 5130 m/s (in iron)
Distance traveled is
d = 100 km = 10⁵ m
Time required to travel in air is
t₁ = d/V₁ = 10⁵/331 = 302.12 s
Time required to travel in iron is
t₂ = d/V₂ = 10⁵/5130 = 19.49 s
The difference in time is
302.12 - 19.49 = 282.63 s
Answer: 283 s (nearest second)
ANSWER: Destructive Interference
-This is the exact definition of the question you have provided, look this term up if you do not believe lol.
Hope this Helps!
Explanation:
The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum