Answer:
Options A, B, and C are all possible.
Explanation:
We know that the instantaneous velocity of the dog at 3:14PM is possitive to toward the flowers. But what about the acceleration to toward the flowers?
If the dog is decreasing speed at 3:14PM, it means that acceleration is negative toward the flowers, hence (since F=ma) the net force points away from the flowers.
If the dog is increasing speed at 3:14PM, it means that acceleration is positive toward the flowers, hence (since F=ma) the net force points toward the flowers.
If the dog is not increasing nor decreasing speed at 3:14PM, it means that acceleration is 0, hence (since F=ma) the net force is null and it does not point neighter to toward the flowers nor away from the flowers. This happens when the forces acting on the dog are equal to both sides.
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
Option(a) the mass of cart 2 is twice that of the mass of cart 1 is the right answer.
The mass of cart 2 is twice that of the mass of cart 1 is correct about the mass of cart 2.
Let's demonstrate the issue using variables:
Let,
m1=mass of cart 1
m2=mass of cart 2
v1 = velocity of cart 1 before collision
v2 = velocity of cart 2 before collision
v' = velocity of the carts after collision
Using the conservation of momentum for perfectly inelastic collisions:
m1v1 + m2v2 = (m1 + m2)v'
v2 = 0 because it is stationary
v' = 1/3*v1
m1v1 = (m1+m2)(1/3)(v1)
m1 = 1/3*m1 + 1/3*m2
1/3*m2 = m1 - 1/3*m1
1/3*m2 = 2/3*m1
m2 = 2m1
From this we can conclude that the mass of cart 2 is twice that of the mass of cart 1.
To learn more about inelastic collision visit:
brainly.com/question/14521843
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