Answer:
P1 = 1.3 (500 + 60) = 728 kg-m total momentum to right at start
P2 = (v2 - 10) 60 + 500 v2
total momentum after running at -10 with respect to cart = 728 where v2 is the new speed of the cart
728 = 560 v2 - 600
v2 = 1328 / 560 = 2.37 m/s new speed of cart
Check:
After: p2 for cart = 500 * 2.37 = 1186
p1 for man = (2.37 - 10) * 60 = -458
P2 = p1 + p2 = 728 total momentum unchanged
PART A) Yes, the fact that there is a frictional force acting on the satellite generates a loss of energy due to friction. What causes satellite to diminish its orbit during its tour. In fact, many satellites have rectifier systems that allow them to position themselves and remain in their orbit for a long time to avoid being trapped by the Earth's gravity Force and fall into the atmosphere where they would probably be torn apart.
PART B) As a similarity, one could start by mentioning the structure of the two equations are similar and have their own constants who were responsible for supporting them. While the law of gravity speaks of the masses of the bodies the electrostatic law speaks of the charges of the bodies. For both the force is inversely proportional to the square of the distance that separates them.
However, the most notable difference between them is basically their statement. While one of the equations speaks about greavedad the other reflects the electromagnetic phenomena. It should be noted that the force of gravity is much weaker than the electromagnetic force and that the latter has the capacity of attraction and repulsion. While the gravitational force only that of attraction.
Answer:

Explanation:
The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:







Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
Answer:
0.1 m
Explanation:
It is given that,
Mass of the object, m = 350 g = 0.35 kg
Spring constant of the spring, k = 5.2 N/m
Amplitude of the oscillation, A = 10 cm = 0.1 m
Frequency of a spring mass system is given by :
Time period: