In this 2-dimensional graph, the x-component of each vector is the horizontal distance from the origin, while the y-component of each vector is the vertical distance from the origin. It can be seen that the c vector is 1 vertical unit away from the origin, which means that it has a y-component of 1.
Answer:
the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.
Explanation:
We can answer this exercise using Gauss's law
Ф = ∫ e . dA = 
/ ε₀
field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field
From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.
Answer:
The squeeze will not regain its shape
Explanation:
The squeeze bottle will not regain its shape.
This is because the atmospheric pressure compresses the squeeze bottle. Since the pressure in the squeeze bottle is now not equal to the atmospheric pressure since it has been corked tightly, its internal pressure cannot balance out the atmospheric pressure and thus cancel its effect.
So, the squeeze bottle does not regain its shape due to this imbalance of pressure.
Answer:
Normal Conversation: i=106i0
i(dB)=60
Power saw a 3 feet: i=1011i0
i(dB)=110
Jet engine at 100 feet: i=1018i0
i(dB)=180
Explanation:
if these are the same as edge, then these are the answers! :)
Answer: 35*10^3 N/m
Explanation: In order to explain this problem we know that the potential energy for spring is given by:
Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.
We also know that with additional streching of 2 cm of teh spring, the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.
Then the difference of energy for both cases is 7 J so:
ΔUp= 1/2*k* (0.02)^2 then
k=2*7/(0.02)^2=35000 N/m
