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il63 [147K]
4 years ago
9

A jogger runs at an average speed of 4.20 mi/h. (a) how fast is she running in m/s? (report your answer to the correct number of

significant figures.) 1.88 m/s (b) how many kilometers does she run in 33.0 min? (report your answer to the correct number of significant figures.) 3.72 km (c) if she starts a run at 11:15
a.m., what time is it after she covers 4.84 × 104 ft? : p.m.
Physics
1 answer:
PolarNik [594]4 years ago
5 0
a) How fast is she running in m / s?
 The first thing you should know for this case is the following conversion
 1mi = 1609.34m
 1h = 3600s
 Applying the conversion
 4.20 (mi / h) * (1609.34 / 1) (m / mi) * (1/3600) (h / s) = 1.88 m / s
 answer
 She is running 1.88 in m / s

 (b) how many kilometers does she run in 33.0 min? 
 By definition, the distance equals the speed by time
 d = v * t
 Also,
 1min = 60seconds
 1 km = 1000 meters.
 We have then
 d = (1.88) * (33) * (60) * (1/1000) = 3.72 Km
 answer
 she run 3.72 Km in 33.0 min
 
(c) if she starts a run at 11:15 a.m., what time is it after she covers 4.84 × 104 ft? : p.m.
 The first thing we need to know is the following conversion
 1Km = 3280.84 feet
 We have then
 3.72 (Km) * (3280.84 / 1) (feet / Km) = 12204.73 feets
 I have the following rule of three we can know how long she travels 4.84 × 104 ft
 12204.73 feets ----> 33 min
 4.84 * 10 ^ 4 feets ----> x
 Clearing x:
 x = ((4.84 * 10 ^ 4) / (12204.73)) * (33) = 131 min
 131 min = 2:11:00 hours
 she starts a run at 11:15 a.m.
 she finishes at
 1:26 p.m.
 answer
  it is 1:26 p.m when she covers 4.84 × 104 ft
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Answer:

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let's use trigonometry

          cos 45 = F₂₄ₓ / F₂₄

          sin 45 = F_{24y) / F₂₄

          F₂₄ₓ = F₂₄ cos 45

          F_{24y} = F₂₄ sin 45

let's do the sum on each axis

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          Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis  

         F_y = - F₂₃ + F_{24y}

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         F₂₁ = F₂₃

Let's use Coulomb's law

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the distance between the two charges is

         r = a

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we calculate F₂₄

           F₂₄ = k q₂ q₄ / r₂₄²

the distance is

           r² = a² + a²

           r² = 2 a²

         

we substitute

           F₂₄ = k  q² / 2 a²

we substitute in the components of the forces

          Fx = - k \frac{q^2}{a^2} +  k \frac{q^2}{2 a^2}  \ cos 45

          Fx = k \frac{q^2}{a^2}  ( -1 + ½ cos 45)

          F_y = k \frac{q^2}{a^2} ( -1 +  ½ sin 45)    

         

We calculate

            F₀ = 9 10⁹ 4.25² / 0.440²

            F₀ = 8.40 10¹¹ N

       

            Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

            Fₓ = -5.43 10¹¹ N

         

remember cos 45 = sin 45

             F_y = - 5.43 10¹¹  N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

          F = -5.43 10¹¹ (i + j)   N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

          F = \sqrt{F_x^2 + F_y^2}

          F = 5.43 10¹¹  √2

          F = 7.68 10¹¹ N

in angle is

           θ = 45º

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Answer:

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b. The load accelerates downward at a rate 0.4 m/s??

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