Answer:
ω = 380π rad/s
Explanation:
The formula for the angular frequency is the oscillation frequency f (hertz) multiplied by 2π
ω = 2πf
then
ω = 2π(190)
ω = 380π rad/s
Answer:
0.0675 seconds
Explanation:
From the question,
We apply newton's second law of motion
F = m(v-u)/t.................... Equation 1
Where F = force exert by the brake, v = final speed, u = initial speed m = mass of the bicycle, t = time.
make t the subject of the equation
t = m(v-u)/F................... Equation 2
Given: m = 180 kg, u = 6.0 m/s, v = 0 m/s (comes to stop), F = -1600 N ( agianst the dirction of motion)
Substitute these value into equation 2
t = 180(0-6.0)/-1600
t = -1080/-1600
t = 0.0675 seconds.
Answer:
Explanation:
Since, Alex is at rest. Therefore, the speed measured by him will be the absolute speed of car P. Therefore, taking easterly direction as positive:
And the absolute velocity of Barbara's Car is given as:
Now, for the velocity of Car p with respect to the velocity of Barbara's Car can be given s follows:
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
The change in velocity is 20 m/s to the north.
