The gravitational force Fg between two objects is given by the equation: Fg=(G*m₁*m₂)/r₂, where G=6.67*10^-11 m³ kg⁻¹ s⁻² is the gravitational constant, m₁ and m₂ are the masses of the two bodies and r is the distance between those bodies. Due to the gravitational attraction the pencil and the eraser will attract if we there is no friction on the surface. m₁=10 g=0.01 kg is the mass of the pencil m₂=20 g=0.02 kg is the mass of the eraser r=2.5 cm = 0.025 m First we calculate the Fg: Fg={(6.67*10^-11)*0.01*0.02}/(0.025²)=2.1344*10^-11 N To get the velocity v of the pencil: v²=2as, where a is the acceleration of the pencil and s is the path. In our case s=r so we can write: v²=2ar a=Fg/m₁= 2.133*10^-9 m/s² v²=2*(2.133*10^-9)*0.025=1.0665*10^-10 v=√(1.0665*10^-10)=1.0327*10^-5 m/s We have the velocity and the acceleration, so we can calculate the time t with the equation: t=v/a=(1.0327*10^-5)/(2.133*10^-9)=4841.6 s 1 hour has 3600 s so when we divide time t in seconds by 3600 we get time T in hours: T=t/3600=4841.6/3600=1.3449 h. So the time for the pencil and eraser to touch is T=1.3449 hours. Also time T can be expressed like T= 1h and 20 mins and 41.64 s

. A pencil (m = 10 g) and eraser (m = 20 g) are placed 2.5 cm apart. If they are left on a surface without friction they will eventually come into contact with one another. Assuming that the pencil is the object to move, how long (in hours) will it take the two objects to touch?

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2016-04-26 05:45:48

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