Answer:
3 ohm
Explanation:
Resistance = Voltage/Current
= 6.0/2
= <u>3</u><u> </u><u>o</u><u>h</u><u>m</u>
Answer:
The first statement is false, the Sun has a stronger gravitational pull.
:
It is TRUE. The pivot point of a lever is called the fulcrum .
<span>net work = change in kinetic energy
for Block B, we just have the force from block A acting on it
F(ab)d= .5(1)vf² - .5(1)(2²)
F(ab)d= .5vf² - 2
Block A, we have the force from the hand going in one direction and the force of block B on A going the opposite direction
10-F(ba)d = .5(4)vf² - .5(4)(2²)
10-F(ba)d = 2vf² - 8
F(ba)d = 18 - 2vf²
now we have two equations:
F(ba)d = 18 - 2vf²
F(ab)d= .5vf² - 2
since the magnitude of F(ba) and F(ab) is the same, substitute and find vf (I already took into account the direction when solving for F(ab)
10-.5vf² + 2 = 2vf² - 8
12 - .5vf² = 2vf² - 8
20 = 2.5vf²
vf² = 8
they both will have the same velocity
KE of block A= .5(4)(2.828²) = 16 J
KE of block B=.5(1)(2.828²) = 4 J</span>
Answer:
a) x = ⅔ d
, b) the charge must be negative, c) Q
Explanation:
a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing
∑ F = 0
-F₁₂ + F₂₃ = 0
F₁₂ = F₂₃
let's replace the values
k Q Q / r₁₂² = k Q 4Q / r₂₃²
Q² / r₁₂² = 4 Q² / r₂₃²
suppose charge 3 is placed at point x
r₁₂ = x
r₂₃ = d-x
we substitute
1 / x² = 4 / (d-x) 2
1 / x = 2 / (d-x)
x = 2 (x-d)
x = 2x -2d
3x = 2d
x = ⅔ d
b) The sign of the charge must be negative, to have an attractive charge on the two initial charges
c) Q
