Answer:
validate
Explanation:
<u>Law of Conservation of energy</u>
According to the law of conservation of energy, energy of a body or a system gets transferred to another system as different energy. We cannot create energy nor we can destroy energy. Energy is always transformed into other forms of energy.
In the context, Joe noticed that the aquarium's temperature dropped during the night and he concluded that according to the law of conservation of energy, the energy of the aquarium in terms of thermal energy is being transferred to the surrounding air.
This is because the aquarium is considered to be an open system where energy can come in and move out.
Hence Joe's statement is valid.
Idkk look it up on quizlet, they flash cards to tell you lots of stuff
Explanation:
the answer is above with its si unit
Answer:
E = 1.19 N/C
Explanation:
Let's first determine the length of the arc which can be given as:
L= Rθ
where:
L = length of the arc
R = radius of curvature
θ = angle in radius
L = (9.09×10⁻²m)(2.59)
L = (0.0909)(2.59)
L = 0.235431 m
Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:
![E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D-sin%28-%5Cfrac%7B%5Ctheta%7D%7B2%7D%29%5D)
![E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%2Bsin%28%5Cfrac%7B%5Ctheta%7D%7B2%7D%29%5D)
![E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D)
Since 
where;
L = length
Q = charge
λ = density of the charge;
then substituting
for λ, we have :
![E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%28%5Cfrac%7BQ%7D%7BL%7D%29%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D)
![E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2Q%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D%7D%7B4%20%5Cpi%20E_oLR%7D)
substituting our given parameter; we have:
![E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%286.26%2A10%5E%7B-12%7DC%29%5Bsin%5Cfrac%7B2.59rad%7D%7B2%7D%5D%7D%7B4%20%5Cpi%20%288.85%2A10%5E%7B-12%7DC%5E2%2FN.m%5E2%29%280.235431%29%280.0909%29%7D)
E = 1.1889 N/C
E = 1.19 N/C
∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C