- #1
Sugi San
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Homework Statement
i need help with this problem (Online HW)
If the force of attraction between a 1035 kg sports car and a 10680 kg cement truck is 3.930×10-6 N, how far are they apart?
Homework Equations
F = Gm_1m_2/r^2
OK.Sugi San said:r^2 = (6.67300 × 10^-11)(1035)(10680) / (3.930 x 10^-6)
= [(6.67300 )(1035)(10680) / (3.930)] x [10^-11/10^-6]
= [(6.67300 )(1035)(10680) / (3.930)] x 10^-5
= 18768958.6 x 10^-5
You are messing up the exponents. When you move the decimal to the left one place you must add +1 to the exponent. You subtracted instead. (Example: 456.7 = 4.567 x 10^2)= 1876895.86 x 10^-6
= 1.87689586 x 10^-12
Once you correctly find r^2, just take the square root.so how we find the distance?
The difference lies in the mass and size of the objects. The truck has a larger mass and size compared to the car, thus creating a stronger gravitational force on other objects.
The gravitational force between the truck and car decreases as the distance between them increases. This is described by the inverse-square law, which states that the force is inversely proportional to the square of the distance between the objects.
Yes, although the force may be very small, gravitational interactions can be observed between any two objects with mass, regardless of their size. However, the effects are more noticeable on a larger scale.
The gravitational force between the truck and car will cause them to accelerate towards each other. This acceleration is dependent on the mass of the objects and the distance between them.
No, there is no limit to the distance at which gravitational interactions can occur. However, as the distance increases, the force becomes weaker and is eventually negligible.