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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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Answer:

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Explanation:

From the question we are told that

    The weight in air is  W_a =  40 \ N

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Now according to Archimedes principle the weight of the object in water is mathematically represented as

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Where  m_w is he mass of the water displaced

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       m_w g  =  40 -20

      m_w g  = 20 \ N --- (1)

Now according to Archimedes principle the weight of the object in unknown  is mathematically represented as

       W_u =  W_a -m _u g

Where  m_u is he mass of the unknown liquid  displaced

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dividing equation 2 by equation 1

      \frac{m_ug}{m_wg}  =  \frac{10}{20}

     \frac{m_u}{m_w}  =  \frac{1}{2}

=>  m_u =  0.5 m_w

Now since the volume of water and liquid displaced are the same then

      \rho _u =  0.5 \rho_w

This because

         density =  \frac{mass}{volume}

So if  volume is constant

         mass = constant * density

Where \rho_u is the density of the liquid

     and  \rho_ w is the density of water which is a constant with a value \rho_w = 1000 kg/m^3

So

        \rho_u  = 1000*0.5

        \rho_u  =500 kg /m^3

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