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Citrus2011 [14]
3 years ago
8

Second and third class levers are differentiated by __________. a. the location of the fulcrum b. the location of the load c. th

e presence of multiple loads d. the type of fulcrum present please select the best answer from the choices provided. a b c d
Physics
1 answer:
strojnjashka [21]3 years ago
5 0

Answer:

the answer is b

Explanation:

Second and third class levers are differentiated by <u>the location of the </u><u>load.</u>

<em>Hope</em><em> </em><em>this</em><em> </em><em>help</em><em> </em><em>you</em><em> </em><em>out </em><em>and have</em><em> </em><em>a </em><em>nice</em><em> </em><em>day </em><em>=</em><em>)</em>

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A 600-kg car accelerates from rest to 10 m/s^2. What is the force on the car?
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Cars cross a certain point on the highway in accordance with a Poisson process with rate = 3 per minute. If Al runs across the h
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Answer:

The probability is 0.2212

Solution:

As per the question:

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X: time of waiting for the next vehicle on the highway

Now,

To find the probability of the next vehicle to arrive after 10 s

The probability distribution function is given by:

f(x) = \lambda e^{- \lambda x}

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P(X < x) = 1 - e^{- \lambda x}

P(X \leq x) = 1 - e^{- 0.05 x}

For X> 0,

P(X > x) = e^{- \lambda x}

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4 0
3 years ago
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
romanna [79]

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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