The wavelength of the third line in the Lyman series, and identify the type of EM radiation
In this series, the spectral lines are obtained when an electron makes a transition from any high energy level (n=2,3,4,5... ). The wavelength of light emitted in this series lies in the ultraviolet region of the electromagnetic spectrum.
1 / lambda = R(h)* ( 
- 
)
= 109678 ( 
- 
)
= 109678 (8/9)
Lambda = 9 / (109678 * 8 )
= 102.6 * 
m = 102.6 nm
To learn more about Lyman series here
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Answer:
it needs to be shaken but make sure you have enough room to shake it safely
Explanation:
To properly operate the laboratory thermometer it needs to be shaken but make sure you have enough room to shake it safely. This done because there is a small bend in the mercury channel of a clinical thermometer that uses mercury. You must shake the thermometer to get the mercury from a previous reading from the thermometer back into the bulb for taking new reading. The bend prevents flow back into the tube so that one can comfortably take reading.
The frequency of a genotype is how often it occurs. This can be obtained by:
number of occurrences / total population
= 160 / 1000
= 0.16 or 16%
Answer:
10.4 m/s
Explanation:
First, find the time it takes for the projectile to fall 6 m.
Given:
y₀ = 6 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.11 s
Now find the horizontal position of the target after that time:
Given:
x₀ = 6 m
v₀ = 5 m/s
a = 0 m/s²
t = 1.11 s
Find: x
x = x₀ + v₀ t + ½ at²
x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²
x = 11.5 m
Finally, find the launch velocity needed to travel that distance in that time.
Given:
x₀ = 0 m
x = 11.5 m
t = 1.11 s
a = 0 m/s²
Find: v₀
(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²
v₀ = 10.4 m/s
Well 200 doubled or (x2)=400 if that’s what it means
