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sasho [114]
3 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]3 years ago
8 0
They will go away from each other
Furkat [3]3 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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Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of 1.39 kW/m2. (a) Assuming that Earth (and
Zigmanuir [339]

Answer:

F=5.8\times 10^{8}\ N

F=35.57\times 10^{21}\ N

Explanation:

Given that

Intensity I

I= 1.39\ KW/m^2

Speed\ of \ light = 2.99\times 10^8\ m/s

Radius of earth,R = 6370 Km

We know that surface area of earth, A

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A=\pi (6370\times 10^3)^2\ m^2

A=1.27\times 10^{14}\ m^2

As we know that pressure due to intensity given as

P=\dfrac{I}{V}

V =Velocity of light

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P=\dfrac{1.39\times 1000}{=3\times 10^8}

P=4.6\times 10^{-6}\ Pa

We know that force F

F = P .A

F=4.6\times 10^{-6}\times 1.27\times 10^{14}\ N

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b)Gravitational force F

F=G\dfrac{m.M}{r^2}

M = mass\ of\ sun = 2\times 10^{30} kg\\m = mass\ of\ earth = 6\times 10^{24}kg

r =1.5\times 10^{11}\ m

G =6.67\times 10^{-11}Nm^2/kg^2

So F

F=6.67\times 10^{-11}\times \dfrac{2\times 10^{30}\times 6\times 10^{24}kg}{1.5\times 10^{11}}

F=35.57\times 10^{21}\ N

7 0
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Answer:

the warm air would be up and cool would be down

Explanation:

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