I believe the answer in Covalent Bond.
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:
E = -4000 N / C
Explanation:
The potential and electric field are related
V = - E s
E = - V / s
we reduce the magnitudes to the SI system
s = 4 mm (1 m / 1000 mm) = 0.004 m
we calculate
E = - 16 /0.004
E = -4000 N / C
Velocity=3.4m/sec
Mass=30kg
so kinetic energy=1/2mv^2
=1/2×30×3.4×3.4
=15×3.4×3.4
=15×11.56
=173.4 kg m per second square
Answer:
7m/s^2
Explanation:
using v=u+at
since the car started from rest, u=0 , v=14m/s t=2s
a =acceleration.
14=0+a×2
14=0+2a
14=2a
a= 14/2 =7
a=7m/s^2