20,000 because I did the quiz and got it right txt me if you need help
The new frequency of oscillation when the car bounces on its springs is 0.447 Hz
<h3>Frequency of oscillation of spring</h3>
The frequency of oscillation of the spring is given by f = (1/2π)√(k/m) where
- k = spring constant and
- m = mass on spring
Now since k is constant, and f ∝ 1/√m.
So, we have f₂/f₁ = √(m₁/m₂) where
- f₁ = initial frequency of spring = 1.0 Hz,
- m₁ = mass of driver,
- f₂ = final frequency of spring and
- m₂ = mass on spring when driver is joined by 4 friends = 5m₁
So, making f₂ subject of the formula, we have
f₂ = [√(m₁/m₂)]f₁
Substituting the values of the variables into the equation, we have
f₂ = [√(m₁/m₂)]f₁
f₂ = [√(m₁/5m₁)]1.0 Hz
f₂ = [√(1/5)]1.0 Hz
f₂ = 1.0 Hz/√5
f₂ = 1.0 Hz/2.236
f₂ = 0.447 Hz
So, the new frequency of oscillation when the car bounces on its springs is 0.447 Hz
Learn more about frequency of oscillation of spring here:
brainly.com/question/15318845
Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
Explanation:
PE = KE
mgh = ½ mv²
v = √(2gh)
v = √(2 × 9.8 m/s² × 0.75 m)
v = 3.83 m/s