Answer:
it is plate tectonics option D hope it helps
there are 7 major and 20 minor plates
Answer:
4.981 MeV
Explanation:
The quantity of energy Q can be calculated using the formula
Q = (mass before - mass after) × c²
Atomic Mass of thorium = 232.038054 u, atomic of Radium = 228.0301069 u and mass of Helium = 4.00260. The difference of atomic number and atomic mass between the thorium and radium ( 232 - 228) and ( 90 - 88) show α particle was emitted.
1 u = 931.494 Mev/c²
Q = (mass before - mass after) × c²
Q = ( mass of thorium - ( mass of Radium + mass of Helium ) )× c²
Q = 232.038054 u - ( 228.0301069 + 4.00260) × c²
Q = 0.0053471 u × c²
replace 1 u = 931.494 MeV/ c²
Q = 0.0053471 × c² × (931.494 MeV / c²)
cancel c² from the equation
Q = 0.0053471 × 931.494 MeV = 4.981 MeV
The problem ask to calculate the bullet's flight time and the bullet's speed as it left the barrel. So base on the problem, the answer would be that the flight time is 0.076 seconds and the speed of the bullet is 657.9 m/s. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications.
Answer:
angular range is ( 0.681 rad , 0.35 rad )
Explanation:
given data
wavelength λ = 380 nm = 380 × 
m
wavelength λ = 700 nm = 700 × m
to find out
angular range of the first-order
solution
we will apply here slit experiment equation that is
d sinθ = m λ ...........1
here m is 1 for single slit and d is = 
so put here value in equation 1 for 380 nm
we get
d sinθ = m λ
sinθ = 1 × 380 ×
θ = 0.35 rad
and for 700 nm
we get
d sinθ = m λ
sinθ = 1 × 700 ×
θ = 0.681 rad
so angular range is ( 0.681 rad , 0.35 rad )
