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maria [59]
3 years ago
11

- The school zone in front of your school has a posted speed limit of 25 mi/h, which is about 11 m/s. Let's

Physics
1 answer:
miv72 [106K]3 years ago
5 0

Answer:

s = 38.7 m

Explanation:

First we calculate the distance covered during uniform motion of reaction time.

s₁ = vt

where,

s₁ = distance covered during uniform motion = ?

v = uniform speed = 11 m/s

t = time = 2.3 s

Therefore,

s₁ = (11 m/s)(2.3 s)

s₁ = 25.3 m

Now, we calculate the distance covered during decelerated motion:

2as₂ = Vf² - Vi²

where,

a = deceleration = -4.5 m/s²

s₂ = distance covered during decelerated motion = ?

Vf = Final Velocity = 0 m/s

Vi = Initial Velocity = 11 m/s

Therefore,

2(-4.5 m/s²)s₂ = (0 m/s)² - (11 m/s)²

s₂ = (-121 m²/s²)/(-9 m/s²)

s₂ = 13.4 m

the total distance will be:

s = s₁ + s₂

s = 25.3 m + 13.4 m

<u>s = 38.7 m</u>

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A 3250-kg aircraft takes 12.5 min to achieve its cruising
scoundrel [369]

Answer:

effeciency n = = 49%

Explanation:

given data:

mass of aircraft 3250 kg

power P = 1500 hp = 1118549.81 watt

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v =85 km/h = 236.11 m/s

n = \frac{P_0}{P}

P_o = \frac{total\ energy}{t} = \frac{ kinetic \energy + gravitational\ energy}{t}

kinetic energy= \frac{1}{2} mv^2  =\frac{1}{2} 3250* 236 = 90590389.66 kg m^2/s^2

kinetic energy = 90590389.66 kg m^2/s^2

gravitational energy = mgh = 3250*9.8*10000 = 315500000.00  kg m^2/s^2

total energy = 90590389.66 +315500000.00 = 409091242.28 kg m^2/s^2

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8 0
3 years ago
The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist
Sidana [21]

Complete Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2

Answer:

The acceleration due to gravity is  g = 167.2 \ m/s^2  

Explanation:

From the question we are told that

     The length of the simple pendulum is L = 1.081.08 \ m

      The number of cycles is  N =  101

       The time take is  t =  2.00 *10^{2 \ }s

Generally the period of this oscillation is mathematically evaluated as

         T = \frac{N}{t }

substituting values

         T = \frac{101}{2.0*10^2 }

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The period of this oscillation is mathematically represented  as

               T = 2 \pi \sqrt{\frac{l}{g} }

making g the subject of the formula we have

              g = \frac{L}{[\frac{T}{2 \pi } ]^2 }

              g = \frac{4 \pi ^2 L }{T^2 }

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               g = \frac{4 * 3.142 ^2  * 1.08 }{505.505^2 }

               g = \frac{4 * 3.142 ^2  * 1.08 }{0.505^2 }  

              g = 167.2 \ m/s^2  

7 0
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