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larisa [96]
2 years ago
9

A car bounces up and down on its springs at 1.0 Hz with only the driver in the car. Now the driver is joined by four friends. Th

e new frequency of oscillation when the car bounces on its springs is
Physics
1 answer:
romanna [79]2 years ago
7 0

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

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Answer:

Explanation:

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T

The box is at a distance A from the point of projection. Then the range R=A

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8 0
4 years ago
The tension in the horizontal cord is 30N. Find the weight of the object.
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Therefore, weight is equal to 30 N. Hope this answers the question. Have a nice day. Feel free to ask more questions.
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One of the equations of gravity is this:
{v}^{2} = {u}^{2} + 2gh
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u = initial velocity which is 0 for objects falling from a height
g = acceleration due to gravity and it is approximately 10m/s^2. It's a constant so pretty much remember this number. It's positive since the work being done is caused by gravity (in other words, it's falling down). It can also be negative if the work being down is against gravity (in other words, it's going up)
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h = \frac{49}{20}
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