Answer: 1666J
Explanation:
Given that,
Mass of box (m) = 10kg
Total distance covered by box (h)
= (5m + 12m)
= 17m
work done on the box = ?
Work is done when force is applied on an object over a distance. Hence, the magnitude of work done on the box depends on its mass (m), distance covered (h), and acceleration due to gravity (g)
(g has a value of 9.8m/s²
i.e Work = mgh
Work = 10kg x 9.8m/s² x 17m
Work = 1666J
Thus, 1666 joules of work was done by the woman on the box.
<span>Answers: (a) 2.0 m/s (b) 4 m/s
Method:
(a) By conservation of momentum, the velocity of the center of mass is unchanged, i.e., 2.0 m/s.
(b) The velocity of the center of mass = (m1v1+m2v2) / (m1+m2)
Since the second mass is initially at rest, vcom = m1v1 / (m1+m2)
Therefore, the initial v1 = vcom (m1+m2) / m1 = 2.0 m/s x 6 = 12 m/s
Since the second mass is initially at rest, v2f = v1i (2m1 /m1+m2 ) = 12 m/s (2/6) = 4 m/s </span>
Answer:
The distance covered is 40 m and the displacement is 31,6m.
Explanation:
The distance covered is the sum of the two distances (10+30). The displacement is equal to the distance of the hipotenusa of the triangle that the two distances (10 m to north and 30m to east) create. Using the Pythagoras theorem the displacent is equal to the Square root of (30^2 +10^2) .
Answer:
The coefficient of static friction between your partner and the floor is 0.55
Explanation:
Given:
Mass 
Kg
Frictional force 
N
From the formula of frictional force,
Where 
coefficient of static friction, 
Put the above values and find the coefficient of static friction.
Therefore, the coefficient of static friction between your partner and the floor is 0.55
