Light strikes a 5.0-cm thick sheet of glass at an angle of incidence in air of 50°. the sheet has parallel faces and the glass h
1 answer:
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Answer:
Explanation:
The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2 i.e. and
The wavelength of Hi line of the Balmer series is given by :
So, the wavelength for this line is 550 nm. Hence, this is the required solution.
Answer:
The best option is for the following option m = 15 [g] and V = 5 [cm³]
Explanation:
We have that the density of a body is defined as the ratio of mass to volume.
where:
Ro = density = 3 [g/cm³]
Now we must determine the densities with each of the given values.
<u>For m = 7 [g] and V = 2.3 [cm³]</u>
<u>For m = 10 [g] and V = 7 [cm³]</u>
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<u>For m = 15 [g] and V = 5 [cm³]</u>
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<u>For m = 21 [g] and V = 8 [cm³]</u>
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Answer:
Fc = [ - 4.45 * 10^-8 j ] N
Explanation:
Given:-
- The masses and the position coordinates from ( 0 , 0 ) are:
Sphere A : ma = 80 kg , ( 0 , 0 )
Sphere B : ma = 60 kg , ( 0.25 , 0 )
Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15
- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2
Find:-
what is the gravitational force on C due to A and B?
Solution:-
- The gravitational force between spheres is given by:
F = G*m1*m2 / r^2
Where, r : The distance between two bodies (sphere).
- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-
Determine the angle (α) between vectors rac and rab using cosine rule:
Determine the angle (β) between vectors rbc and rab using cosine rule:
- Now determine the scalar gravitational forces due to sphere A and B on C:
Between sphere A and C:
Fac = G*ma*mc / rac^2
Fac = (6.674×10−11)*80*0.2 / 0.2^2
Fac = 2.67*10^-8 N
vector Fac = Fac* [ - cos (α) i + - sin (α) j ]
vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]
vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N
Between sphere B and C:
Fbc = G*mb*mc / rbc^2
Fbc = (6.674×10−11)*60*0.2 / 0.15^2
Fbc = 3.56*10^-8 N
vector Fbc = Fbc* [ cos (β) i - sin (β) j ]
vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]
vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N
- The Net gravitational force can now be determined from vector additon of Fac and Fbc:
Fc = vector Fac + vector Fbc
Fc = [ - 2.136 i - 1.602 j ]*10^-8 + [ 2.136 i - 2.848 j ]*10^-8
Fc = [ - 4.45 * 10^-8 j ] N