Determine the slit spacing d. Explain which measurement you made, show your calculation and your result for the slit spacing. Th
1 answer:
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Answer:
The string must support the tension of 392 N.
Explanation:
The tension that the string must support should equal the centripetal force exerted on the on the stone as it goes in a circular path (because if the string supported less tension, it would break).
The centripetal force exerted on the stone is
where
<em>v</em> = velocity of the stone in m/s
<em>m</em> = mass of the stone in kg
<em>R</em> = radius of the circular path.
Now the velocity of the stone is 7.00 m/s, the mass of the stone is 4000g or 4 kg (1000 g = 1kg), and the radius of the circular path is just the length of the string, and it is 50 cm or 0.5 m (100cm =1m); therefore, we get
m = 4kg
v =7m/s
R = 0.5m.
We put these values into the equation for the centripetal force and get:
The centripetal force is 392 Newtons, and therefore, the tension that the string must support mus be 392 N.
Answer:
The object would weight 63 N on the Earth surface
Explanation:
We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:
Now, if the body is on the surface of the Earth, its weight (w) would be:
Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion: