Answer:
C 2000v its obviously ans because if o is 1000 2 vo is 2000v
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
Utilize the formula: 
= Final Velocity (86 m/s)
= Initial Velocity (0 m/s)
a = acceleration (m/s²)
t = Time (100 seconds)
As a result,
86 m/s = 0 + (a)(100 seconds)
Using algebra, divide 86 m/s by 100 seconds:
86 m/s = 100a
a = 0.86 m/s²
Rounded to one decimal place: 0.9 m/s²
Let me know if you have any questions!
Answer: yea ma’am I’m sorry but you still
Answer:
371.2 mm
Explanation:
The Balmer series of spectral lines is obtained from the formula
1/λ = R(1/2² -1/n²) where λ = wavelength, R = Rydberg's constant = 1.097 × 10⁷ m⁻¹
when n = 15
1/λ = 1.097 × 10⁷ m⁻¹(1/2² -1/15²)
= 1.097 × 10⁷ m⁻¹(1/4 -1/225)
= 1.097 × 10⁷ m⁻¹(0.25 - 0.0044)
= 1.097 × 10⁷ m⁻¹ 0.245556
= 2.693 10⁶ m⁻¹
So,
λ = 1/2.693 10⁶ m⁻¹
= 0.3712 10⁻⁶ m
= 371.2 mm
