please help with 1 and 2
Acceleration due to gravity g-9.8 m/s Universal gravitational constant, G-6.67 x 10 Mass of Earth 5.98 x 10 kg Radius of Earth 6.38x10 km N.m/kg átteupt All questions- circle the richt answer In the figure below, the weight of the block is 80.0 N. The horizontal pulling force on the bl is 0.40. Th block is ock is 50.0 N. The coefficient of kinetic friction between the block and the surface e block was pulled a distance 5.0 m. The work done by the net force on the F-50.0 N (A) 15003 (8) 240.0J ( 250J (D) 90.03 ()75.03 (F None of these 2. The force applied parallel to the x-axis by a child on a box of mass 10.0 kg is shown in the figure below. The work done by the force on the box from x-0tox 8.0 m is 10 (A) 12.0J (B) 20.0J (C) 44.0J (D) 32.0J (E) 80.0J (F) None of the above
Solution
here,
1)
weight of block , W1 = 80 N
F = 50 N
uk = 0.4 , s = 5 m
the work done , W = net force .s
W = (F - uk * W1) . 5
W = ( 50 - 0.4 * 80) * 5 J
W = 90 J
the work done by net force is C) 25 J
2)
mass , m = 10 kg
the work done , W = area under the curve
W = 0.5 * (4 -0) * ( 10 - 0) + (8-4) * (6 -0) J
W = 44 J
the work done is C) 44 J
.
Measures of Dispersion and Variability: Range, QD, AD and SD
please help with 1 and 2 Acceleration due to gravity g-9-8 m-s Univers.docx
1. please help with 1 and 2
Acceleration due to gravity g-9.8 m/s Universal gravitational constant, G-6.67 x 10 Mass of
Earth 5.98 x 10 kg Radius of Earth 6.38x10 km N.m/kg átteupt All questions- circle the richt
answer In the figure below, the weight of the block is 80.0 N. The horizontal pulling force on the
bl is 0.40. Th block is ock is 50.0 N. The coefficient of kinetic friction between the block and the
surface e block was pulled a distance 5.0 m. The work done by the net force on the F-50.0 N (A)
15003 (8) 240.0J ( 250J (D) 90.03 ()75.03 (F None of these 2. The force applied parallel to the x-
axis by a child on a box of mass 10.0 kg is shown in the figure below. The work done by the
force on the box from x-0tox 8.0 m is 10 (A) 12.0J (B) 20.0J (C) 44.0J (D) 32.0J (E) 80.0J (F)
None of the above
Solution
here,
1)
weight of block , W1 = 80 N
F = 50 N
uk = 0.4 , s = 5 m
the work done , W = net force .s
W = (F - uk * W1) . 5
W = ( 50 - 0.4 * 80) * 5 J
W = 90 J
the work done by net force is C) 25 J
2)
2. mass , m = 10 kg
the work done , W = area under the curve
W = 0.5 * (4 -0) * ( 10 - 0) + (8-4) * (6 -0) J
W = 44 J
the work done is C) 44 J
