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sasho [114]
4 years ago
12

Two identical soccer balls are rolled towards each other. What will be true after they collide head-on

Physics
2 answers:
tresset_1 [31]4 years ago
8 0
They will go away from each other
Furkat [3]4 years ago
6 0

Answer:

(v_{i_{1} }+v_{i_{2} }) = (v_{f_{1} }+v_{f_{2} })

Explanation:

There are some things we can say:

  • If the Collision was Elastic, then the conservation of energy applies, where the kinetic energy of the system before they collide will be the same after they collide.
  • If the collision was inelastic, then we can assert that the kinetic energy of the system before the collision and after the collision will be different, because there is some energy lost in the form of heat for example, though the energy of the entire system stays the same (if we add the losses).
  • The law of conservation of momentum states that the momentum of the system before the collision is exactly the same after the collision so if m_{1} =m_{2}:

m_{i_{1} } *v_{i_{1} }+m_{i_{2} } *v_{i_{2} } = m_{f_{1} } *v_{f_{1} }+m_{f_{2} } *v_{f_{2} }

m_{i_{1} } *(v_{i_{1} }+v_{i_{2} }) = m_{f_{1} } *(v_{f_{1} }+v_{f_{2} })

Since there is conservation of mass as well (we assume that nothing breaks apart after the collision):

(v_{i_{1} }+v_{i_{2} }) = (v_{f_{1} }+v_{f_{2} })

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A bird flying straight upward at 5 m/s drops a berry when it is 300 m above the ground. How fast is the berry going when it hits
Nikitich [7]

Answer:

v=77.62 m/s

Explanation:

Given that

h= - 300 m

speed of the bird ,u= 5 m/s

Lets take Speed of the berry when it hit the ground = v m/s

we know that ,if object is moving upward

v² = u² - 2 g h

u=Initial speed

v=Final speed

h=Height

Now by putting the values

v² = u² - 2 g h

v² = 5² - 2 x 10 x (-300)                ( take g = 10 m/s²)

v² =25 + 20 x 300

v² ==25 + 6000

v² =6025

v=77.62 m/s

Therefore the final speed of the berry will be 77.62 m/s.

5 0
3 years ago
Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate
masha68 [24]

Answer:

The heat loss per unit length is   \frac{Q}{L}   = 2981 W/m

Explanation:

From the question we are told that

     The outer diameter of the pipe is d = 104mm = \frac{104}{1000} = 0.104 m

     The thickness is  D = 2mm = \frac{2}{1000} = 0.002m  

      The temperature  of water is  T = 90^oC = 90 + 273 = 363K  

      The outside air temperature is T_a = -10^oC = -10 +273 = 263K

        The water side heat transfer coefficient is z_1 = 300 W/ m^2 \cdot K

       The  heat transfer coefficient is  z_2 = 20 W/m^2 \cdot K

The heat lost per unit length is mathematically represented as

           \frac{Q}{L}   = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1}  +  \frac{ln [\frac{d}{D} ]}{z_2}}

Substituting values

         \frac{Q}{L}   = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300}  +  \frac{ln [\frac{0.104}{0.002} ]}{20}}

           \frac{Q}{L}   = \frac{628}{0.2107}

           \frac{Q}{L}   = 2981 W/m

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