(amount of heat)Q = ? , (Mass) m= 4 g , ΔT = T f - T i = 180 c° - 20 °c = 160 °c ,
Ce = 0.093 cal/g. °c
Q = m C ΔT
Q = 4 g × 0.093 cal/g.c° × ( 180 °c- 20 °c )
Q= 4×0.093 × 160
Q = 59.52 cal
I hope I helped you^_^
Answer:
The contribution of the wavelets lying on the back of the wave front is zero because of something known as the Obliquity Factor. It is assumed that the amplitude of the secondary wavelets is not independent of the direction of propagation, Sources: byju's.com
Answer:
Any Lens
Explanation:
I Hope it's right if not so Sorry :)
Answer: 4.0024 x 10^ -11 m or 0.040024 nm
Explanation:
λ = h c/ΔE
λ = wave lenght
h = 6.626 x 10 ^ -34 m² kg /s = planck constant
ΔE = 31 keV potential ( 1 keV = 1.6021 x 10^-16J)
c = velocity of light = 3 x 10⁸ m/s
substitute gives
λ = <u>6.626 x 10 ^ -34 m² kg /s x 3 x 10⁸ m/s</u> = 4.0024 x 10^ -11 m
31 x 1.6021x10^-16 J
