Answer:
The answers are in the explanation section below
Explanation:
1) The generalization that can be made from the exploration is that as we move away from the positive electrode, the potential energy gets lower. If we move away from the negative electrode, then the potential energy becomes higher.
2) The positive test charge will have the least potential energy when it gets to the negative electrode point.
3) To move one electron 1m in a direction along one of the equal potential lines, there is no energy needed. Zero work will be required for a charge to move on the equipotential line.
4) If lightning strikes a tree 20m away, it would be better to face the tree or have our back facing the tree. This is because the equipotential line will be present at the point where our body stands, this will protect from electric shock.
The pattern to be sketched is attached.
Answer:
0.03 A
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12 V
Resistor (R) = 470 Ω
Current (I) =?
From ohm's law, the voltage, current and resistor are related by the following formula:
Voltage = current × resistor
V = IR
With the above formula, we can obtain the current in the circuit as follow:
Voltage (V) = 12 V
Resistor (R) = 470 Ω
Current (I) =?
V = IR
12 = I × 470
Divide both side by 470
I = 12 / 470
I = 0.03 A
Thus, the current in the circuit is 0.03 A
Responder:
A) ω = 565.56 rad / seg
B) f = 90Hz
C) 0.011111s
Explicación:
Dado que:
Velocidad = 5400 rpm (revolución por minuto)
La velocidad angular (ω) = 2πf
Donde f = frecuencia
ω = 5400 rev / minuto
1 minuto = 60 segundos
2πrad = I revolución
Por lo tanto,
ω = 5400 * (rev / min) * (1 min / 60s) * (2πrad / 1 rev)
ω = (5400 * 2πrad) / 60 s
ω = 10800πrad / 60 s
ω = 180πrad / seg
ω = 565.56 rad / seg
SI)
Dado que :
ω = 2πf
donde f = frecuencia, ω = velocidad angular en rad / s
f = ω / 2π
f = 565.56 / 2π
f = 90.011669
f = 90 Hz
C) Periodo (T)
Recordar T = 1 / f
Por lo tanto,
T = 1/90
T = 0.0111111s
<span>You should deflect the
ball in order to maximize your speed on the skateboard.
Since this creates a larger impulse, you want to deflect the ball. Splitting it
up into catching and throwing the ball may by something you can think of deflecting
the ball. First, you need to catch the ball, which in turn would push you
forward with some speed. (The speed we are talking about should obviously be
equal to option A, where you catch the ball). Now, throw the ball back to him
since these two processes are equal to deflecting the ball. Throwing a mass away
from you would cause or enable you to move even fast.</span>
