A 2.50-kg textbook is forced against a horizontal spring of negligible mass and force constant 250 N/m, compressing the spring a
1 answer:
Answer:
L = 1.06 m
Explanation:
According to the work-energy theorem, the total work performed by all forces must match the difference in kinetic energy.
This leads us to:
Next, we know that
Since both the initial and final speeds of the book are zero, its kinetic energy at both points is also zero.
Given that
k = 250 N/m
x = 0.250 m
m = 2.50 kg
Now we will insert all the known values into the equation.
At this point, we find
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Answer:
The correct choice is C: points 1, 4, and 5 are equal, followed by 2 and 3 being equal.
Explanation:
Here’s the breakdown:
The electric field from the positive sheets E₁ = б/2E₀
E₂ is from the negative sheet = -б/2E₀
At points 1, 4, and 5, the electric fields created by the sheets oppose each other.
At point 1, the total field is calculated as -E₁ + E₂ = 0.
Similarly, at point A, the total field results in -E₁ - E₂ = 0.
However, at any point in between the plates, the electric field is directed consistently in one way.
At points 2 and 3, the field is directed to the right.
Thus, we have:
E net = E₁ + E₂
= б/2E₀ + -б/2E₀
=б/E₀
Note: Please refer to the attached document for the full question accompanying this solution.
Answer:
The final size is nearly the same as the initial size because the increase in size is remarkably small
Solution:
According to the problem:
The proton beam energy is E = 250 GeV =
Distance traveled by the photon, d = 1 km = 1000 m
Proton mass,
Initial size of the wave packet,
Now,
This operates under relativistic principles
The rest mass energy for the proton is expressed as:
This proton energy is
Thus, the speed of the proton, v
The time to cover 1 km = 1000 m of distance is calculated as:
T =
T =
According to the dispersion factor;
Thus, the widening of the wave packet is relatively minor.
Hence, we can conclude that:
where
= final width
Answer:
Explanation:
The position of the charge q₁ is established at (0,0)
Meanwhile, the charge q₂ is located at (x₁,0)
Thus, the electric potential energy between these two charges is determined by:
Now, the location of charge q₂ shifts from (x₁,0) to (x₂,y₂). The updated electric potential energy between the charges can be represented as:
According to the work-energy theorem, the alteration in potential energy corresponds to the work performed. This is expressed mathematically as:
Consequently, the work done by the electrostatic force on the moving charge is . Therefore, this concludes the solution.