C. unbalanced is the correct answer for newton's first law.
The angular acceleration of a rotating object is given by
where
is the final angular speed of the object
is its initial angular speed
is the time taken to accelerate
For the wheel in our problem,
,
and
, so its angular acceleration is
I see the light moving exactly at speed equal to c.
In fact, the second postulate of special relativity states that:
"The speed of light in free space has the same value c<span> in all inertial frames of reference."
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The problem says that I am moving at speed 2/3 c, so my motion is a uniform motion (constant speed). This means I am in an inertial frame of reference, so the speed of light in this frame must be equal to c.
Answer:
Explanation:
As we know that the length of the conductor is given as
now if it is converted into a square then we have
now the are of the loop will be
now the magnetic flux is defined as
here we know
B = 1.0 T
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
