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Lynna [10]
3 years ago
15

An automobile traveling along a straight road increases its speed from 72 ft/s to 84 ft/s in 180 ft. if the acceleration is cons

tant, how much time elapses while the auto moves the 180 ft? answer in units of s.
Physics
1 answer:
Nikolay [14]3 years ago
5 0
The equation that would allow us to calculate for the acceleration given the distance is written below,

      a = (Vf² - Vo²) / 2d

where a is the acceleration, Vf is the final velocity, Vo is the initial velocity, and d is distance. 

Substituting the known values,
    a = ((84 ft/s)² - (72 ft/s)²) / 2(180 ft) = 5.2 ft/s²

Then, the equation that would relate the initial velocity, distance, acceleration and time is calculated through the equation,
      
     d = Vot + 0.5at²

Substituting the known values,
    180 = 72(t) + 0.5(5.2)(t²)

The value of t from the equation is 2.3 s

<em>ANSWER: 2.3 s</em>
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1.Predict the frequency of a tuning fork that emits a sound with a wavelength of 0.385 m.
lina2011 [118]

Answer:

1.) Frequency F = 890.9 Hz

2.) Wavelength (λ) = 0.893 m

Explanation:

1.) Given that the wavelength = 0.385m

The speed of sound = 343 m / s

To predict the frequency, let us use the formula V = F λ

Where (λ) = wavelength = 0.385m

343 = F × 0.385

F = 343/0.385

F = 890.9 Hz

2.) Given that the frequency = 384Hz

Using the formula again

V = F λ

λ = V/F

Wavelength (λ) = 343/384

Wavelength (λ) = 0.893 m

The two questions can be solved with the use of formula

3 0
2 years ago
A mobile starts with a speed of 250m / s and begins to decelerate at a rate of 3m / s². How fast is it after 45s?
Korvikt [17]

\large{ \underline{ \underline{ \bf{ \purple{Given}}}}}

  • Speed of the mobile = 250 m/s
  • It starts decelerating at a rate of 3 m/s²
  • Time travelled = 45s

\large{ \underline{ \underline{ \bf{ \green{To \: find}}}}}

  • Velocity of mobile after 45 seconds

\large{ \underline{ \underline{ \red{ \bf{Now, \: What \: to \: do?}}}}}

We can solve the above question using the three equations of motion which are:-

  • v = u + at
  • s = ut + 1/2 at²
  • v² = u² + 2as

So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

\large{ \bf{ \underline{ \underline{ \orange{Solution:}}}}}

We are provided with,

  • u = 250 m/s
  • a = -3 m/s²
  • t = 45 s

By using 1st equation of motion,

⇛ v = u + at

⇛ v = 250 + (-3)45

⇛ v = 250 - 135 m/s

⇛ v = 115 m/s

✤ <u>Final</u><u> </u><u>velocity</u><u> </u><u>of</u><u> </u><u>mobile</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u>5</u><u> </u><u>m</u><u>/</u><u>s</u>

<u>━━━━━━━━━━━━━━━━━━━━</u>

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How will the motion of the arrow change after it leaves the bow?
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The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.

pls thank me and brainliest me

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