Answer:
20.7
Explanation:
:0 because basis of the daily occured
I would say that work is being done in all the situations because technically in physics work is done whenever force is applied through a distance.
Answer:
(a) Both the charges are positive or negative.
(b) Teh value of each charge is 1.53 x 10^-5 C.
Explanation:
Spring constant, K = 340 N/m
Natural length, L = 0.4 m
stretch, y = 0.033 m
(a) Let the charge on each sphere is q and they repel each other so the nature of charge of either sphere may be both positive or both negative.
(b) The electrostatic force is balanced by the spring force.
Answer:
1) The Newton's law the Hulk is most closely demonstrating is <em>the Third Law of motion</em>
Newton's third law of motion states that action and reaction are equal and opposite.
2) The reason why the Newton's Law the Hulk is most closely demonstrating is the Newton's Third Law of Motion is that according to Newton's Third Law of motion, forces exist in pairs, the action of the Hulk when he leaps upward by pushing against the the Earth (which can be assumed stationary in relation to the Hulk), is equal to the reaction of the Earth, which moves down slightly, away from its initial position.
However, due to the large mass of the Earth, compared to the mass of the Hulk, the downward motion of the Earth due to the reaction force (equal to the force with which the Hulk leaps) is negligibly small, such that the Earth can absorb the Hulk's leap force by reacting mainly locally, at the leap point by forming a crater, while the Earth in general, remains in the same place
Explanation:
Apply the Pythagorean theorem to get the resultant velocity:
V = 
Given values:
Vx = 57.8m/s
Vy = 5.6m/s
Plug in and solve for V:
V = 58.1m/s
EDIT: Let's get the direction of the resultant velocity as well.
This equation will give the angle of the velocity as measured off of the ground:
θ = tan⁻¹(Vy/Vx)
Again, the given values are:
Vx = 57.8m/s
Vy = 5.6m/s
Plug in the values and solve for the angle θ:
θ = tan⁻¹(5.6/57.8)
θ = 5.5°
The resultant velocity is oriented 5.5° off the ground.
