7. 1 points Giancol6 5 P035 MyNotes Ask Your What is the distance from the Earth

Question

7. 1 points Giancol6 5 P035 MyNotes Ask Your What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/24 of its value at the Earth's surface? -5 points Gancoi6 5P053. My Notes Ask Your What will a spring scale read for the weight of a 50 kg woman in an elevator that moves as follows? (a) upward with constant speed of 5.0 m/s (b) downward with constant speed of 5.0 m/s (c) upward with acceleration of 0.38g (d) downward with acceleration 0.38g (e) in free fall 9 My Notes Ask Your -2 points Ganool6 5 P 016. A bucket of mass 1.60 kg is whirled in a vertical circle of radius 1.50 m. At the lowest point of its motion the tension in the rope supporting the bucket e is 25.0 N (a) Find the speed of the bucket. m/s (b) How fast must the bucket move at the top of the circle so that the rope does not go slack?

Explanation / Answer

7) We know that outside the Earth's sphere, g is inversely proportional to the distance squared from the center of Earth.

we use g(0) to refer to a point on the Earth's surface and g(1) to the unknown point we need to calculate:

we have

g(1)/g(0) = { r(0) / r(1) }2

g(1)/g(0) = r(0)2 / r(1)2

or  

r(1)2 = r(0)2 * { g(0)/g(1) }

r(1) = r(0) * sqrt{ g(0)/g(1) }

taking r(0) as 6400 km (radius of earth) and g(0)/g(1) as 24

r(1) = 6400 * sqrt (24) km

or r(1) = 31353.47 km

8) for constant speed, the weight will remain same as normal

thus

a) W = mg = 50*9.8 = 490 N

b) W = mg = 50*9.8 = 490 N

c) while acclerating upward with an accleration of 0.38g, we have

W = m*(g + 0.38g) = 50(1.38g) = 50*1.38*9.8 = 676.2 N

d) for acclerating downwards with an accleration of 0.38g, we have

W = m*(g - 0.38g) = 50(0.62g) = 50*0.62*9.8 = 303.8 N

e) for free fall, the scale is also acclerating down at same rate as the woman, thus

W = m*(g-g) = 0

9) At lowest point of its motion , the tension in the rope supporting the bucket is given by

T = W + F = 25.0 N.

(a) We have T = 25 = mg + F = 1.60*9.8 + F = 15.68 + F

we know that

F = mv2/R = 25 - 15.68

v = sqrt((25 - 15.68)*1.5/1.6) = sqrt(8.74) = 2.96 m/s.

(b) For this condition

g = a = u2/R

we get u = sqrt(Rg) = sqrt(1.5*9.8) = sqrt(14.7) = 3.83 m/s.

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.