One of Kepler's laws is that the orbits of planets are elliptical. It's not a suggestion.
BTW, circles are ellipses too, but so special that their likelihood is close to zero.
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
brainly.com/question/5762197
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Answer:
Explanation:
Given that:
R = (0.1) m
To find the electric field for r < R by using Gauss Law
For r < R
where;
Answer:
Energy consumed by the electric kettle in 9.5 min =Pt=(2.5×10
3
)×(9.5×60)=14.25×10
5
J
Energy usefully consumed =msΔT=3×(4.2×10
3
)×(100−15)=10.71×10
5
where s=4.2J/g
o
C= specific heat of water and boiling point temp=100
o
C
Heat lost =14.25×10
5
−10.71×10
5
=3.54×10
5
Answer:
Explanation:
a )
While breaking initial velocity u = 62.5 mph
= 62.5 x 1760 x 3 / (60 x 60 ) ft /s
= 91.66 ft / s
distance trvelled s = 150 ft
v² = u² - 2as
0 = 91.66² - 2 a x 150
a = - 28 ft / s²
b ) While accelerating initial velocity u = 0
distance travelled s = .24 mi
time = 19.3 s
s = ut + 1/2 at²
s is distance travelled in time t with acceleration a ,
.24 = 0 + 1/2 a x 19.3²
a = .001288 mi/s²
= 2.06 m /s²
c )
If distance travelled s = .25 mi
final velocity v = ? a = .001288 mi / s²
v² = u² + 2as
= 0 + 2 x .001288 x .25
= .000644
v = .025 mi / s
= .0025 x 60 x 60 mi / h
= 91.35 mph .
d ) initial velocity u = 59 mph
= 86.53 ft / s
final velocity = 0
acceleration = - 28 ft /s²
v = u - at
0 = 86.53 - 28 t
t = 3 sec approx .
