In a parallel circuit, both bulbs are connected directly to the power supply
or battery. As long as the power supply is capable of running both of them,
neither bulb is aware of the other's presence. If one bulb fails 'open', the
other bulb doesn't change at all. The voltage across it, the current through it,
its resistance, and its brightness,all remain exactly as they were before.
Answer:
¿Es una pregunta de la escuela o de la vida?
Explanation:
The answer would be C.30J
Answer:
a. V3 = 19.36m/s
b. β3 = -18.8°
Explanation:
Our given data is:
mt = 6.4kg; Vo = [10.2*cos(35°), 10.2*sin(35°)] m/s
m1 = 1.4kg; V1 = [-3.7, 0] m/s
m2 = 1.8kg; V2 = [0, 11.1] m/s
By inspection:
m3 = 3.2kg V3 = [ V3x, V3y]
At the point where the ball explodes, there's only x-component of the velocity, so:
Pix = Pfx and Piy = Pfy
On x-axis:
mt * Vo*cos(35°) = m1 * V1x + m3 * V3x
Solving for V3x:
V3x = 18.33 m/s
On y-axis:
0 = m2 * V2y + m3 * V3y
Solving for V3y:
V3y = -6.24 m/s
Therefore, the magnitude is:
![V3 = \sqrt{V3x^2 + V3y^2} = 19.36m/s](https://tex.z-dn.net/?f=V3%20%3D%20%5Csqrt%7BV3x%5E2%20%2B%20V3y%5E2%7D%20%3D%2019.36m%2Fs)
The angle is:
![\beta 3 = atan(\frac{V3y}{V3x}) = -18.8\°](https://tex.z-dn.net/?f=%5Cbeta%203%20%3D%20atan%28%5Cfrac%7BV3y%7D%7BV3x%7D%29%20%3D%20-18.8%5C%C2%B0)
it is known as the
nuclear meltdown