Answer:
Jorge wanted to know where the apples were
Explanation:
Ya see, I don't want to explain, cuz I guessed. Also I'm on episode 499 of naruto so yeah
Answer:
The voltage drop across the bulb is 115 V
Explanation:
The voltage drop equation is given by:

Where:
ΔW is the total work done (4.6kJ)
Δq is the total charge
We need to use the definition of electric current to find Δq

Where:
I is the current (2 A)
Δt is the time (20 s)


Then, we can put this value of charge in the voltage equation.

Therefore, the voltage drop across the bulb is 115 V.
I hope it helps you!
All of Dina's potential energy Ep is converted into kinetic energy Ek so Ep=Ek, where Ep=m*g*h and Ek=(1/2)*m*v². m is the mass of Dina, h is the height of ski slope, g=9.8 m/s² and v is the maximal velocity.
So we solve for v:
m*g*h=(1/2)*m*v², masses cancel out,
g*h=(1/2)*v², we multiply by 2,
2*g*h=v² and take the square root to get v
√(2*g*h)=v, we plug in the numbers and get:
v=9.9 m/s.
So Dina's maximum velocity on the bottom of the ski slope is v=9.9 m/s.
Answer:
the height reached is = 0.458 [m]
Explanation:
We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:
![Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Cwhere%3AEk%3D%20kinetic%20energy%20%5BJ%5D%5C%5Cm%20%3D%20mass%20of%20the%20ball%20%5Bkg%5D%5C%5Cv%20%3D%20velocity%20of%20the%20ball%20%5Bm%2Fs%5D)
Replacing the values on the equation we have:
![Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%2A%282%29%2A%283%5E%7B2%7D%20%29%5C%5C%20Ek%3D9%5BJ%5D%5C%5C)
This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.
![Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\](https://tex.z-dn.net/?f=Ek%3DEp%5C%5Cwhere%5C%5CEp%3Dpotential%20energy%20%5BJ%5D%5C%5CEp%3Dm%2Ag%2Ah%5C%5Cwhere%5C%5Cg%3Dgravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Ch%3Dheight%20reached%20%5Bm%5D%5C%5C)
Now we have:
![h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]](https://tex.z-dn.net/?f=h%3D%5Cfrac%7BEp%7D%7Bm%2Ag%7D%20%5C%5Ch%3D%5Cfrac%7B9%7D%7B2%2A9.81%7D%20%5C%5C%5C%5Ch%3D0.45%20%5Bm%5D)
In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]