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miyahhh123
07.03.2020 •
Physics
Problem 1: Problem 9.24 in Young & Freedman An electric turntable 0.750 m in diameter is rotating about a fixed axis with an initial angular velocity of 0.250 rev/s and a constant angular acceleration of 0.900 rev /s 2 . (a) Compute the angular velocity of the turntable after 0.200 s. (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at = 0.200 s? (d) What is the magnitude of the resultant acceleration of a point on the rim at = 0.200 s?
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Ответ:
x₁ = 345100 km
Explanation:
The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:
Let call "x₁" distance between center of the earth and the object, and
"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object
we can express:
F₁ ( force between earth and the object )
F₁ = K * Mt * m₀/ ( x₁)² K is a gravitational constant
F₂ (force between mn and the object)
F₂ = K * Ml * m₀ / (x₂)²
Then:
F₁ = F₂ K*Mt*m₀ / x₁² = K*Ml*m₀ /x₂²
Or simplifying the expression
Mt/ x₁² = Ml/ x₂²
We know that x₁ + x₂ = 384000 Km then
x₁ = 384000 - x₂
Mt/( 384000 - x₂)² = Ml / x₂²
Mt * x₂² = Ml *( 384000 - x₂)²
We need to solve for x₂
Mt * x₂² = Ml *[ ( 384000)² + x₂² - 768000*x₂]
By substitution:
5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -
7.348*10∧22*768000*x₂
Simplifying by 10∧22
5.972*10²*x₂² = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂
Sorting out
5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂
(597,2 - 7,348 )* x₂² = 10.80*10∧11 - 56.43*10∧5*x₂
590x₂² + 56.43*10∧5*x₂ - 10.80*10∧11 = 0
Is a second degree equation
x₂ = -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11 / 1160
x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11 / 1160
x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 254880*10∧10 / 1160
x₂ ₁ = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160
x₂ ₁ = -56.43*10∧5 + 508* 10∧5 / 1160
x₂ ₁ = 451.27*10∧5/1160
x₂ ₁ = 4512.7*10∧4 /1160
x₂ ₁ = 3.89*10∧4 km (distance between the moon and the object)
x₂ ₁ = 38900 km
x₂ = 38900 km
We dismiss the other solution because is negative and there is not a negative distance
Then the distance between the earth and the object is:
x₁ = 384000 - x₂
x₁ = 384000 - 38900
x₁ = 345100 km