A runner is jogging in a straight line at a steady vr= 7.3 km/hr. When the runner is L= 2.1 km from the finish line, a bird begins flying straight from the runner to the finish line at vb= 29.2 km/hr (4 times as fast as the runner). When the bird reaches the finish line, it turns around and flies directly back to the runner After this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back to the runner. The bird repeats the back and forth trips until the runner reaches the finish line. How far does the bird travel from the beginning (including the distance traveled to the first encounter

Answers

Answer 1

Answer:

Explanation:

Time taken by jogger to travel the distance to finishing line = 2.1 / 7.3

= .28767 hr

Bird will keep flying for this time period

distance covered by bird = speed x time

= 29.2 x .28767 km

= 8.4 km .


Related Questions

Mamie Clark was a psychologist who is known for her research on __________. A. intellectual and developmental disabilities and gifted children B. racism and its adverse effects on child development C. identity development among Asians living in the US D. academic development of children in desegregated schools

Answers

Answer:

B. racism and its adverse effects on child development

Explanation:

Dr. Mamie Clark studied the effects of segregation and racism on the self-esteem of black children.

According to Clark in O'Connell & Russo, 2001, p. 271; "To separate them from others of similar age and qualifications solely because of their race generates a feeling of inferiority as to their status in the community that may affect their hearts and minds in a way unlikely ever to be undone...."

One of her works titled "The Development of Consciousness of Self and the Emergence of Racial Identification in Negro Preschool Children" investigated the early level of conscious racial identity in black preschool children.

Answer:

B

Explanation:

cuz its B

A cannonball with a mass of 50 kilograms and a golf ball with a mass of 0. 8 kilograms are dropped in a vacuum from a height of 10 meters. Compare the acceleration of each object as they fall.

Answers

It's gonna take a huge amount of time and effort, and we're gonna spend a huge amount of money, to build a vacuum chamber big enough to do this experiment.  

So when we're finally ready to try it, the cannonball and golf ball will make a  kind of cute experiment, but let's make it really interesting.  Let's ALSO try it with a feather, an apple, a watermelon, a tennis ball, a basketball, a coffeepot, a chair, an old computer, a set of dishes, a trombone, a bicycle, a math book, a TV set, a gallon of milk, a skateboard, a shotgun, a concrete block, a car, a waterbed, a schoolbus, a battleship, your dog, and my wife !

Drop every single one of them from a height of 10 meters.

Every single one of them accelerates at 9.8 m/s² as it falls, takes 1.429 seconds to reach the floor, and hits the floor at a speed of 14 m/s.  The feather, the schoolbus, and my wife stay together all the way down.

The only way this doesn't happen is if ...

-- you do this experiment in some other place that's not Earth, or

-- you throw one of the objects and you don't just drop it, or

-- the vacuum chamber has a leak and some air gets into it.

Because that's how gravity works.

The level of toluene (a flammable hydrocarbon) in a storage tank may fluctuate between 10 and 400 cm from the top of the tank. since it is impossible to see inside the tank, an open-end manometer with water or mercury as the manometer fluid is to be used to determine the toluene level. one leg of the manometer is attached to the tank 500 cm from the top. a nitrogen blanket at atmospheric pressure is maintained over the tank contents. felder, richard m.; rousseau, ronald w.; bullard, lisa g.. elementary principles of chemical processes, 4th edition (page 81). wiley. kindle edition.

Answers

Complete Question  

The complete question is shown on the first and second uploaded image  

Answer:

When water is used the reading is  [tex]    R =  2281.6 \  cm  [/tex]

When mercury is used the reading is [tex]   R =  23.83 \ cm [/tex]  

The best fluid to use is mercury because for water a slight change in toluene level will cause a  large change in height .

Explanation:

From the question we are told that  

    The length of the leg of the manometer to the top of the tank is  d =  500cm

   The toluene level where in the tank where the height of the manometer fluid level in the open arm is equal to the height where the manometer is connected to the tank is  h =150 cm  

   The manometer reading is  R  

Generally at the point where the height of the open arm is equal to the height of the of the point connected to the tank ,  

   The pressure at the height of the both arms of the manometer corresponding to the base of the tank are equal  

     i.e   [tex]P_1 = P_2[/tex]

Here [tex]P_1[/tex] is the pressure of the manometer at the point corresponding to the base of the tank and this is mathematically represented as  

       [tex] P_{atm} + P_1 =  P_{atm} + P_t[/tex]

Here [tex]P_t[/tex] is the pressure due to the toluene level in the tank and in the arm of the manometer connected to the tank and this is mathematically represented as  

         [tex]P_t  =  \rho_t  * g  * h_i[/tex]

Here  

      [tex]\rho_t [/tex] is the density of toluene with value  [tex]\rho_t =  867 kg/m^3 [/tex]

       

 [tex]h_i[/tex] is the height of the connected arm above the point equivalent to the base of the tank , this mathematically represented as

          [tex]h_i =  d - h + R[/tex]

  and  [tex] P_2 [/tex] is the the pressure at the open arm of the manometer at the point equivalent to the base of the base of the tank and this is mathematically represented as

         [tex] P_2 =  \rho_f * g *  h_f [/tex]

Here  

       [tex]\rho_f[/tex] is the density of the fluid in use , if it is water the density is  

       [tex]\rho_w =  1000 \  kg /m^3 [/tex]

and  if it is  mercury the density is  

        [tex]\rho_m =  13600 \  kg /m^3 [/tex]

[tex]h_f[/tex] is the height of the  fluid in the open arm of the manometer from the point equivalent to the base of the tank which is equivalent the manometer reading R

So when the fluid is water we have

      [tex] P_{atm} +  \rho_t* g *(d - h + R) =  P_{atm} + \rho_f * g *  h_f[/tex]

=>   [tex]   \rho_t* (d - h + R) =   \rho_w *  h_f[/tex]        

=>   [tex]    867  (500 - 150 + R) =    1000 *  R [/tex]

=>    [tex]    R =  2281.6 \  cm  [/tex]

So when the fluid is mercury we have      

  [tex]   \rho_t* (d - h + R) =   \rho_m *  h_f[/tex]          

=>   [tex]    867  (500 - 150 + R) =   13600  *  R [/tex]  

=>   [tex]   R =  23.83 \ cm [/tex]  

The difference in the mercury reading for mercury due to the fact that they have different densities as we have seen in this calculation

So the best fluid to use is mercury because for water a slight change in toluene level will cause a  large change in height .

       

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