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Komok [63]
4 years ago
13

A bowling ball (mass = 7.2 kg, radius = 0.15 m) and a billiard ball (mass = 0.31 kg, radius = 0.028 m) may each be treated as un

iform spheres. what is the magnitude of the maximum gravitational force that each can exert on the other?
Physics
1 answer:
loris [4]4 years ago
6 0
<span>4.7x10^-9 newtons Since both the billiard ball and the bowling ball are uniform spheres, you can simplify the math by assuming that both are gravitational point sources centered within each sphere. And since the gravitational attraction increases as the bodies move closer together, you want to place them as close as possible to each other which would be touching. So when touching, the distance between the two point sources of gravity would be the sum of their radiuses. 0.15m + 0.028m = 0.178m The formula for gravitational attraction is F = G(m1m2)/(r^2) G = 6.67x10^-11 m^3/(s^2 kg) m1, m2 = masses r = radius. So substituting the known values gives F = 6.67x10^-11 m^3/(s^2 kg) (7.2 kg * 0.31 kg) / (0.178 m^2) F = 6.67x10^-11 m^3/(s^2 kg) (2.232 kg^2) / (0.031684 m^2) F = 6.67x10^-11 m/(s^2) (2.232 kg) / (0.031684) F = 6.67x10^-11 m/(s^2) 70.44565 kg F = 4.69872x10^-9 kg m/s^2 F = 4.69872x10^-9 newtons Since the number of significant figures provided is only 2, the result should be rounded to 2 significant digits giving a result of 4.7x10^-9 newtons</span>
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The tension in the first and second rope are; 147 Newton and 98 Newton respectively.

Given the data in the question

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To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

F = m\ *\ a

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity  g = 9.8m/s^2 )

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Total mass hanging on it; m_T = m_1 + m_2 = 5.0kg + 10.0kg = 15.0kg

So Tension of the rope;

F = m\ * \ g\\\\F = 15.0kg \ * 9.8m/s^2\\\\F = 147 kg.m/s^2\\\\F = 147N

Therefore, the tension in the first rope is 147 Newton

For the Second Rope

Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

F = m\ * \ g\\\\F = 10.0kg \ * 9.8m/s^2\\\\F = 98 kg.m/s^2\\\\F = 98N

Therefore, the tension in the second rope is 98 Newton

Learn More, brainly.com/question/18288215

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