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il63 [147K]
3 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]3 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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Explanation:

Speed of the marathon runner, v = 9.51 mi/hr

Distance covered by the runner, d = 26.220 mile

Let t is the time taken by the marathon runner. We know that the speed of the runner is given by total distance divided by total time taken. Mathematically, it is given by :

v=\dfrac{d}{t}

t=\dfrac{d}{v}

t=\dfrac{26.220\ mi}{9.51\ mi/hr}

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3 years ago
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Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂  : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F  : Force ( N)

d  : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the  bar

mm = 1.5 kg :  mass of the  monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey  (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances  from forces the point 2

d₁₂ = (3-0.6) m = 2.4m  : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium  of moments at the point  2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

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Answer:

F = - K x

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f = 2 π ω = 6.28 * 10.1 / sec = 63.5 / sec

P = 1/f = .0157 sec

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