Respuesta :
Answer:
(a). The 2.0 g metal cube move fast than the 4 g metal cube.
(b). The time taken is 10.19 sec.
Explanation:
Given that,
Mass of metal cube = 2.0 g
Mass of second metal cube = 4.0 g
Coefficient of static friction = 0.65
Charging rate = 7.0 nC/s
Distance = 6.0 cm
(a). We need to calculate the frictional force for 2.0 g metal cube
Using formula of frictional force
[tex]F_{1}=\mu\ mg[/tex]
Put the value into the formula
[tex]F_{1}=0.65\times2.0\times10^{-3}\times9.8[/tex]
[tex]F_{1}=12.74\times10^{-3}\ N[/tex]
The frictional force for 4.0 g metal cube
[tex]F_{2}=0.65\times4.0\times10^{-3}\times9.8[/tex]
[tex]F_{2}=25.48\times10^{-3}\ N[/tex]
The motion of the cube is less when the friction is more.
The 2.0 g metal cube is having less frictional force.
Therefore, the 2.0 g metal cube move fast than the 4 g metal cube.
(b). We need to calculate the time
Firstly we need to calculate the charge
Using electrostatic force and frictional force
[tex]F=\mu\ mg[/tex]
[tex]\dfrac{kq^2}{r^2}=12.74\times10^{-3}[/tex]
Put the value into the formula
[tex]9\times10^{9}\times\dfrac{q^2}{(6.0\times10^{-2})^2}=12.74\times10^{-3}[/tex]
[tex]q^2=\dfrac{12.74\times10^{-3}\times(6.0\times10^{-2})^2}{9\times10^{9}}[/tex]
[tex]q^2=\sqrt{5.096\times10^{-15}}[/tex]
[tex]q=7.138\times10^{-8}\ C[/tex]
The time taken after charging beings is
[tex]t=\dfrac{q}{C}[/tex]
Put the value into the formula
[tex]t=\dfrac{7.138\times10^{-8}}{7.0\times10^{-9}}[/tex]
[tex]t=10.19\ sec[/tex]
Hence, (a). The 2.0 g metal cube move fast than the 4 g metal cube.
(b). The time taken is 10.19 sec.
