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2 years ago

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The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob

ject will be dropped with no initial speed. The objects are initially 179 cm above the ground and separated by 142 cm. What will be the difference in the landing locations of the two objects?

1 answer:

Crazy boy [7]2 years ago

4 0

Answer:

104.3 cm or 179.7

Explanation:

First find time that it takes for the object to hit the ground

$\sqrt{(2H)/g} -> \sqrt{(2 x 179)/ 9.8} = 6.04s\\$ *

Then find xf of projectile $xf= 5.9(6.04) = 37.7\\\\$

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7 or 142+37.7

hope that helps

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Answer:

The value is $E_i = 1.5596 *10^{-18} \ J$

Explanation:

From the question we are told that

The wavelength is $\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m$

The velocity is $v = 2.371*10^6 \ m/s$

The mass of electron is $m_e = 9.109*10^{-31} \ kg$

Generally the energy of the incident light is mathematically represented as

$E = \frac{h * c}{\lambda}$

Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

h is the Planck constant with value $h = 6.62607015 * 10^{-34 } J\cdot s$

So

$E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}$

=> $E = 4.12 *10^{-18} \ J$

Generally the kinetic energy is mathematically represented as

$E_k = \frac{1}{2} * m_e * v^2$

=> $E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2$

=> $E_k = 2.56 *0^{-18} \ J$

Generally the ionization energy is mathematically represented as

$E_i = 4.12 *10^{-18} - 2.56 *0^{-18}$

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3 years ago

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When you go right into detail, the fundamental laws governing these proofs are very technical. They have differential equations to show as proof. It is too detailed to discuss here. The important things is that, these fundamental laws are what explains the science in our basic activities and natural phenomena:

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