Estimate the force required to bind the two protons in the He nucleus together. (Hint: Model the protons as point charges. Assum
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Answer: hello below is the missing part of your question
A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.
answer
x = 0.0962 m
Explanation:
<em>First step :</em>
Determine the length of the rough patch/spot
F = Uₓ (mg)
and w = F.d = Uₓ (mg) * d
hence;
d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m
<em>next : </em>
work done on unstretched spring length
Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m
w' = Uₓ (mg) * d
= 0.49 * 10 * 9.81 * 0.4847 = 23.27 J
also given that the Elastic energy of spring = work done ( w')
1/2 * kx^2 = 23.27 J
x = = 0.0962 m
Answer:
Va is two times greater than Vb
Explanation:
The maximum height reached by the balls are:
Since we are told that Ya = 4Yb:
Simplifying the equation:
Answer:
Displacement =
Magnitude of displacement = 13.89 units
Angle of displacement = 30.26°
Explanation:
Mark the co-ordinates of each of the end points of the vectors.
The marked co-ordinates are shown below.
Displacement is the shortest distance between two points. Here, displacement between the starting and end points is the vector .
Displacement vector for a point A from origin O is given as:
Magnitude is given as:
Direction of the vector is given as:
Therefore, displacement vector is:
Magnitude of displacement is given as:
Angle of displacement is:
°
The change in internal energy of a system is given by (second law of thermodynamics)
where Q is the heat absorbed by the system and W is the work done on the system.
In order to correctly evaluate the internal energy change, we must be careful with the signs of Q and W:
Q positive -> Q absorbed by the system
Q negative -> Q released by the system
W positive -> W done on the system by the surroundings
W negative -> W done by the system on the surroundings
In our problem, the heat released by the system is
(with negative sign since it is released by the system), and the work done is 
still with negative sign because it is performed by the system on the surrounding, so the change in internal energy is
where Q is the heat absorbed by the system and W is the work done on the system.
In order to correctly evaluate the internal energy change, we must be careful with the signs of Q and W:
Q positive -> Q absorbed by the system
Q negative -> Q released by the system
W positive -> W done on the system by the surroundings
W negative -> W done by the system on the surroundings
In our problem, the heat released by the system is