Rohit kumar

 when an object move from one place to another place with the respect of time             :

                       Is called motion.

                                  Distance :

1. it is the total path covered by a moving object.
2. It is denoted by ‘s’ and its unit is metre.
3. It is scalar quantity.

                               Speed: 

1. speed is the distance covered by an object per unit time .
2. Speed is denoted by ‘v’
3. s.i unit of speed is m/s
4. it is scalar quantity. And speed =distance /time

                       Displacement :

1.  it is the shortest distance between the final point to intinal point .
2. it is the vector quantity .
3. s.i unit of displacement =m/s.

                         velocity :

1. it is defined as displacement per unit time.
2. velocity is denoted by “V”
3. velocity is a vector quantity
4. s.i unit of velocity is m/s

                     acceleration :

1. it is defined as the rate of change of velocity.
2. Acceleration  =v-u/t
3. Si unit of acceleration is m/s2

                       Retardation:

1.  retardation is the change in velocity per unit time when the intinal velocity is more than the final velocity.
2. s.i unit is m/s2

                       equation of motion

1. v=u+at
2. s=ut+1/2at2
3. v2=u2-2as1) Define the scalar quantity and name 5 physical quantities which are scalar and give reasons why these physical quantities are scalar?
   ANS- Scalar quantity is the physical quantity in which only magnitude is measured and                       direction is not specified .
             Eg- Time, Speed, Volume, Temperature, Area.
                     All these quantities are scalar quantities because there direction is not specified, their only                                 magnitude is measured
   
   
   24-JUNE-2016
   
   
   2) GRAPH
       
   

   

   
   

   
   

   
   

   Graph of the above table:-

   
   

   
   
   8
              In  m s–1  = 13.9 m s–1.
   
   
   3 . Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to         the other and back along the same straight path. Find the average speed and average velocity of           Usha. 
   ANS- Total distance covered by Usha in 1 min is 180 m.
              Displacement of Usha in 1 min = 0 m
              Average speed of the object =    180 = 3 m s^-1
                                                                                        ⁶⁰
                
             ^{Average velocity = Displacement   =  0m = 0} m s^-1
                                           Total Time          60
   
   
   4.Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s–1 in 30 s.    Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s.      Calculate the acceleration of the bicycle in both the cases.
   ANS- In the first case 
             U=0 , V=6 m s-1 , t = 30sec , a = ?
             a = V-U   =  6-0 =  1  m s-2  =  0.2 m s–2
                      t           30     5 
   
   
   In the second case 
             U=6 m s-1 , V=4 m s-1 , t = 5sec , a = ?
             a = V-U   =  4-6 =  -2  m s-2  =  -0.4 m s–2
                      t             5       5 
   
   
   5. A train starting from rest attains a velocity of 72 km h–1 in 5 minutes. Assuming that the                     acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for               attaining this velocity.
   ANS- U=0, V= 72 km h-1 = 20 m s-1, t = 5min = 300sec
        (i) a = V-U   =  20-0 =  1  m s-2 
                      t           300    15  
        (ii) by IInd equation of motion
              S= ut + 1/2 at*2
              S= 0X300 + 1/2 X 1/15 X 90000
              S= 3000m = 3km
   
   
   6. A car accelerates uniformly from 18 km h–1 to 36 km h–1 in 5 s. Calculate 
   (i) the acceleration and 
   (ii) the distance covered by the car in that time.
   ANS- U= 18 km h-1 = 5 m s-1 , V= 36 km h-1 = 10 m s-1 , t = 5 s  
        (i) a = V-U   =  10-5 =  1  m s-2 
                      t            5          
        (ii) by II nd equation of motion
              S= ut + 1/2 at*2
              S= 5X5 + 1/2 X 1 X 25
              S= 25 + 12.5 = 37.5 m
   
   
   7. The brakes applied to a car produce an acceleration of 6 m s-2 in the opposite direction to the             motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels         during this time.
   ANS- a = -6 m s-2 , t = 2 sec , V= 0 m s-2 
             by I st equation of motion 
             V=U+at
             0= U+(-12)
             0=U-12
             U= 12 m s-1
              
            by II nd equation of motion
            S= ut + 1/2 at*2
            S= 12X2 + 1/2 X -6 X 4
            S= 24 - 12 = 12 m   IMPROVEMENT IN FOOD RESOURCES
     

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   These nutrients include minerals and vitamins. Unlike macro nutrients, these are required in very minute amounts. Together, they are extremely important for the normal functioning of the body. Their main function is to enable the many chemical reactions to occur in the body. Nevertheless micro nutrients do not function for the provision of energy.
   
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4.