Given: μS = 0.4, m = 0.25 kg, g = 9.8 m/s2
To find: Force (FL)
Formula: FL = μSN = μS(mg)
Calculation: From formula,
FL = 0.4 × 0.25 × 9.8 = 0.98 N
The horizontal force applied...
`24x^(2)-6xy+29y+6x-58y-151=0`
`2(x-3y+3)^(2)+2(3x+y-1)^(2)=180`
or `((x-3y+3)^(2))/(60)+((3x+y-1)^(2))/(90)=1`
or `((x-3y+3)/(sqrt(1+3^(2))sqrt(6)))^(2)+((3x+y-1)/(3sqrt(1+3^(2))))=1`
Thus, C is an ellipse whose lengths of axes are `6,2sqrt(6)`.
The minor and the major axes are `x-3y+3=0 and 3x+y-1=0`, respectively.
Their point...