The distance mirror M2 must be moved so that one wavelength has produced one more new maxima than the other wavelength is;
<u><em>L = 57.88 mm</em></u>
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We are given;
Wavelength 1; λ₁ = 589 nm = 589 × 10⁻⁹ m
Wavelength 2; λ₂ = 589.6 nm = 589.6 × 10⁻⁹ m
We are told that L₁ = L₂. Thus, we will adopt L.
Formula for the number of bright fringe shift is;
m = 2L/λ
Thus;
For Wavelength 1;
m₁ = 2L/(589 × 10⁻⁹)
For wavelength 2;
m₂ = 2L/(589.6)
Now, we are told that one wavelength must have produced one more new maxima than the other wavelength. Thus;
m₁ - m₂ = 2
Plugging in the values of m₁ and m₂ gives;
(2L/589) - (2L/589.6) = 2
divide through by 2 to get;
L[(1/589) - (1/589.6)] = 1
L(1.728 × 10⁻⁶) = 1
L = 1/(1.728 × 10⁻⁶)
L = 578790.67 nm
L = 57.88 mm
Read more at; brainly.com/question/17161594
Answer:
3.62m/s and 2.83m/s
Explanation:
Apply conservation of momentum
For vertical component,
Pfy = Piy
m* Vof (sin38) - m*Vgf (sin52) = 0
Divide through by m
Vof(sin38) - Vgf(sin52) = 0
Vof(sin38) = Vgf(sin52)
Vof (sin38/sin52) = Vgf
0.7813Vof = Vgf
For horizontal component
Pxf= Pxi
m* Vof (cos38) - m*Vgf (cos52) = m*4.6
Divide through by m
Vof(cos38) + Vgf(cos52) = 4.6
Recall that
0.7813Vof = Vgf
Vof(cos38) + 0.7813 Vof(cos52) = 4.6
0.7880Vof + 0.4810Vof = 4.
1.269Vof = 4.6
Vof = 4.6/1.269
Vof = 3.62m/s
Recall that
0.7813Vof = Vgf
Vgf = 0.7813 * 3.62
Vgf = 2.83m/s
(a) +9.30 kg m/s
The impulse exerted on an object is equal to its change in momentum:

where
m is the mass of the object
is the change in velocity of the object, with
v = final velocity
u = initial velocity
For the volleyball in this problem:
m = 0.272 kg
u = -12.6 m/s
v = +21.6 m/s
So the impulse is

(b) 155 N
The impulse can also be rewritten as

where
F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)
is the duration of the collision
In this situation, we have

So we can re-arrange the equation to find the magnitude of the average force:
