A conducting bar moves along a circuit with a constant velocity. A constant magnetic field is perpendicular to the bar and circu
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Answer:
a) r₁ = 3.836 10⁷ m, b) F = - 3,390 10⁸ N , c) R = 120.3 m
Explanation:
a) This is a problem of equilibrium where the force is gravitational, we call F1 the force of the Moon and F2 the force from the earth.
F₁ -F₂ = 0
F₁ = F₂
G m / r₁² = G m
/ r₂²
Let's look for the measured distance from a Coordinate System located on the Moon,
r₂ = D - r₁
Where D is the average distance from Terra to the Moon 3.84 10⁸ m
/ r₁² =
/ (D - r₁)²
(D² - 2 D r₁ + r₁²) = /
r₁²
(1 - /
)r₁² - 2D r₁ + D² = 0
Let's replace and solve the second degree equation
(1 - 5.98 10²⁴ / 7.36 10²²) r₁² - 2 3.84 10⁸ r₁ + (3.84 10⁸)² = 0
-80.25 r₁² - 7.68 10⁸ r₁ + 14.75 10¹⁶ = 0
r₁² + 9.57 10⁶ r₁ - 1.838 10¹⁵ = 0
r₁ = [-9.57 10⁶ ±√ (91.58 10¹² + 7.352 10¹⁵)] / 2
r₁ = [-9.57 10⁶ + - 86.28 10⁶] / 2
The results are:
r₁’= 38.355 10 6 m
r₁ ’’ = -47.915 10 6 m
We take the positive result that a distance between the moon and the Earth, the equalization point is r₁ = 3.836 10⁷ m
b) The force at point R = 2 r₁
R = 2 3.8355 10⁷ = 7.671 10⁷ m
F = F₁ - F₂
F = G m / R² - G m
/ (D- R)²
F = G m ( / R² -
/ (D-R)²)
F = m 6.67 10⁻¹¹ (7.36 10²² / (7.671 10⁷)² - 5.98 10²⁴ / (3.84 10⁸ - 0.7671 10⁸)²
F = m 6.67 10⁻¹¹ (0.125076 10⁸ - 0.63329 10⁸)
F = m (-3.3897 10⁸) N
The mass m of the rocket must be known, suppose it is worth 1 kg (m = 1 kg)
F = - 3,390 10⁸ N
c) let's use gravitational force from the moon
F = G m / R²
R =√ G m / F
R = √ (6.67 10⁻¹¹ 1 7.36 10²² / 3.3897 10⁸)
R = √ (1.4482 10⁴)
R = 1.2034 10² m
R = 120.3 m