Answer : D) Circuit A
This circuit is the only circuit where it is not complete, having and open spot towards the bottom of it, making it and open circuit.
Answer:<em> Option (D) is correct.</em>
Explanation:
Considering the hypothesis elaborated in this comprehension, it's given that areas that tend to lie near forest fires usually have extra positive strikes since smoke carries positively charged particles. In rudimentary term, this states that occurrence of positively charged particles will result in extra positive strikes.
Option (D) states that occurrence of extra positive strikes will be there even weeks after the charge of smoke particles have been dissociated.
Therefore this option, most seriously undermines the hypothesis.
The answer is A and B
Explanation: the nucleus was viewed as composed of combinations of protons and electrons, the two elementary particles known at the time, but that model presented several experimental and theoretical contradictions.
AND
Protons and neutrons have approximately the same mass, about 1.67 × 10-24 grams.
Answer:
A) 4037.2[km]; B) 21472[m/s]
Explanation:
A)
This part can be solved using the principle of energy conservation, where kinetic energy will be equal to potential energy. We will define the kinetic energy and the potential energy.
![Ek= 0.5*m*v^2\\where:\\m = mass [kg]\\v = 8.90*10^3[m/s]\\Ep=m*g*h\\where:\\g = gravity = 9.81[m/s^2]\\h = elevation or height [m]\\](https://tex.z-dn.net/?f=Ek%3D%200.5%2Am%2Av%5E2%5C%5Cwhere%3A%5C%5Cm%20%3D%20mass%20%5Bkg%5D%5C%5Cv%20%3D%208.90%2A10%5E3%5Bm%2Fs%5D%5C%5CEp%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cg%20%3D%20gravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Ch%20%3D%20elevation%20or%20height%20%5Bm%5D%5C%5C)
Now we have to match both equations in this way the mass value is canceled and we can clear h.
![0.5*m*(8.9*10^3)^{2}=m*9.81*h\\ h=4037206.93[m] = 4037.2[km]](https://tex.z-dn.net/?f=0.5%2Am%2A%288.9%2A10%5E3%29%5E%7B2%7D%3Dm%2A9.81%2Ah%5C%5C%20h%3D4037206.93%5Bm%5D%20%3D%204037.2%5Bkm%5D)
B)
To solve this problem we can use the kinematic equations, but first we must identify the initial data:
yo = 2.35*10^7[m]
y = 0 [m] when hit the ground [m]
vo = 0 [m/s], It is essentially at rest...
![v_{f} ^{2} =v_{o} ^{2} +2*a*(y)\\v_{f} =\sqrt{2*9.81*2.35*10^{7} } \\v_{f} =21472.54[m/s] o 21.47[km/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%20%3Dv_%7Bo%7D%20%5E%7B2%7D%20%2B2%2Aa%2A%28y%29%5C%5Cv_%7Bf%7D%20%3D%5Csqrt%7B2%2A9.81%2A2.35%2A10%5E%7B7%7D%20%7D%20%5C%5Cv_%7Bf%7D%20%3D21472.54%5Bm%2Fs%5D%20o%2021.47%5Bkm%2Fs%5D)
It’s C “ Ac can be transported over long distances.”
