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Nuetrik [128]
3 years ago
13

define momentum of a body.state law of co?servation of momentum.prove with the help of third law of motion that total momentum o

f two bodies is conserved during collision provided no external force acts.
Physics
1 answer:
xxTIMURxx [149]3 years ago
4 0
The momentum of a body is defined as the product of its mass and velocity....

law of momentum of conservation depends on collision,
for a collision occurring between object 1 and object 2 in an isolated system,the total momentum of the two objects before the Collision is equal to the total momentum of two objects after the Collision.....

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A tuna fish body is streamlined to reduce friction. Which describes the type of friction the fish must overcome?
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A Cessna aircraft has a lift off speed of 147 km/h. What minimum constant acceleration does this require if the aircraft is to b
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Answer:

The acceleration is  a =51945 \ km/h^2

Explanation:

From the question we are told that

   The lift up speed is  v  = 147 \  km/h

    The distance covered for the take off run is s =  208 m = 0.208 \ km

Generally from kinematic equation we have that

      v^2 = u^2 + 2as

Here u is the initial  speed of the aircraft with value 0 m/ s give that the aircraft started from rest

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=>  a =51945 \ km/h^2

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2 years ago
Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19
Alisiya [41]

Answer:

The workdone is  W = 9.28 * 10^{3} J

Explanation:

From the question we are told that

   The height of the cylinder is  h = 0.588\ m

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    The density of the cylinder is \rho  =  0.346 * \rho_w

     Where \rho_w is the density of freshwater which has a constant value

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Now  

     Let the final height of the device under the water be  =  h_f

      Let  the initial volume underwater be = V_n

     Let the initial height under water be  = h_i

      Let the final volume under water be  = V_f

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Now the work done is mathematically represented as  

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               =   \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.

              = \frac{g A \rho}{2}  [h^2 - h_f^2]

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Substituting values

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        W = 9.28 * 10^{3} J

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