A force vector F1 points due
east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The
resultant of the two vectors has a magnitude of 400 newtons and points along
the due east/west line. Find the magnitude and direction of F2. Note that there
are two answers.
<span>The given values are
F1 = 200 N</span>
F2 =?
Total = 400 N
Solution:
F1 + F2 = T
200 N + F2 = 400N
F2 = 400 - 200
F2 = 200
N
Answer:
x = D (M/M-m) 2.41
Explanation:
a) Let's apply Newton's second law to find the summation of force, where each force is given by the law of universal gravitation
F = g m₁m₂ / r²
Σ F = 0
F1- F2 = 0
F1 = F2
We set the reference system in the body of greatest mass (M) the planet
F1 = g m₁ M / x²
F2 = G m1 m / (D-x)²
G m₁ M / x² = G m₁ m / (D-x)²
M (D-x)² = m x²
MD² -2MD x + M x² = m x²
x² (M-m) -2MD x + MD² = 0
We solve the second degree equation
x = [2MD ±√ (4M²D² - 4 (M-m) MD²)] / 2 (M-m)
x = {2MD ± 2D √ (M² + (M-m) M)} / 2 (M-m)
x = D {M ± Ra (2M²-mM)} / (M-m)
x = D (M ± M √ (2-m/M)) / (M-m)
x = D (M / (M-m)) (1 ±√ (2-m/M)
Let's analyze this result, the value of M-m >> 1, so if we take the negative root, the value of x would be negative, it is out of the point between the two bodies, so the correct result must be taken with the positive root
x = D (M / (M-m)) (1 + √2)
x = D (M/M-m) 2.41
b) X = 2/3 D
x = D (M/M-m) 2.41
2/3 D = D (M/(M-m)) 2.41
2/3 (M-m) = M 2.41
2/3 M - 2/3 m = 2.41 M
1.743 M = 0.667 m
M/m = 0.667/1.743
M/m = 0.38
Answer:
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Answer:
1. The standard metric unit of momentum is the kg•m/s.
2. the units of momentum will be the product of the units of mass and velocity. Mass is measured in kg and velocity in ms-1, therefore, the SI unit of momentum will be kgm/s(-1).
3.Recall that acceleration is rate of change of velocity, so we can rewrite the Second Law: force = mass x rate of change of velocity. Now, the momentum is mv, mass x velocity. ... rate of change of momentum = mass x rate of change of velocity.
Explanation:
i really hope i helped sorry for the paragraphs ;( !
