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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:

The correct answer is;

Change of phase.

Explanation:

When the physical state in which a substance exists in nature changes to another physical state such as a change from solid state to a liquid stat or from a liquid state to a gaseous state, the substance is said to have undergone a phase change, or phase transformation.

The state of a substance will change, depending on the surrounding temperature and the applied pressure and the energy required for the substance to change its state is known as latent heat.

The observed plateaus are periods the latent heat is absorbed to completely change the phase of the substance.

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3 years ago
Henry can lift a 200 N load 20 m up a ladder in 40 s. Ricardo can lift twice the load up one-half the distance in the same amoun
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Henry will lift 200 N load 20 m up a ladder in 40 s.  While the Ricardo will take 400 N load in 80 seconds. So, For Henry to take 400 N load it will take him 80 seconds in two attempts. And,also, he will have to cover 40 m of distance. 
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In a nuclear fusion reaction, the mass of the products is _____ the mass of the reactants.
JulijaS [17]

Answer:

more than

Explanation:

In a nuclear fusion reaction, the mass of the products is more than the mass of the reactants.

7 0
3 years ago
Bill is farsighted and has a near point located 121 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm fro
Arada [10]

Answer:

Explanation:

The lens equation is

1 / f = 1 / di + 1 / do

Where

f is focal length

di is the image distance

do is the object distance

Both Annie and Billy use a glass whose near point is 25cm

Then, the object distance is

do = 25 - 2 = 23cm

The have the same object distance.

Let find the vocal length of bills eye

Given that,

Bill near point is 121cm and distance of the glass from the eye is 2cm

Then,

Image distance of bill is

di_B = -(121-2) = -119cm

object distance do = 23cm

Then,

1 / f_B = 1 / di_B + 1 / do

1 / f_B = -1 / 119 + 1 / 23

1 / f_B = -119^-1 + 23^-1

1 / f_B = 0.0351

Then, f_B = 28.51 cm

Also, let find Annie focal length

Given that,

Annie near point is 74 cm and distance of the glass from the eye is 2cm

Then,

Image distance of Annie is

di_A = -(74-2) = -72cm

object distance do = 23cm

Then,

1 / f_A = 1 / di_A + 1 / do

1 / f_A = -1 / 72 + 1 / 23

1 / f_A = -72^-1 + 23^-1

1 / f_A = 0.02959

Then, f_A = 33.8 cm

Distance of object from the lens when Annie uses Billy glass

Then,

1 / f_B = 1 / di_A + 1 / do

1 / 28.51 = -1 / 72 + 1 / do

28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

do^-1 = 0.048964

do = 20.42 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_A = 20.42 + 2 = 22.42cm

do_A = 22.42 cm

Distance of object from the lens when Billy uses Annie glass

Then,

1 / f_A = 1 / di_B + 1 / do

1 / 33.8 = -1 / 119 + 1 / do

33.8^-1 = -119^-1 + do^-1

do^-1 = 33.8^-1 + 119^-1

do^-1 = 0.03799

do = 26.32 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_B = 26.32 + 2 = 28.32 cm

do_B = 28.32 cm

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Which of the following is NOT common of elite Shang burials?
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Answer:

Large above ground mausoleums were not common in the elite Shang burials.

Explanation:

Large, above the ground mausoleums were not common so the answer is option B.

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