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luda_lava [24]
3 years ago
11

A projectile is fired at v 0 = 381.0 m/s at an angle of θ = 73.5 ∘ , with respect to the horizontal. Assume that air friction wi

ll shorten the range by 34.1 % . How far will the projectile travel in the horizontal direction, R ?
Physics
1 answer:
Ipatiy [6.2K]3 years ago
5 0

Answer:

Explanation:

velocity of projection, vo = 381 m/s

angle of projection, θ = 73.5°

The formula for the range is

R=\frac{u^{2}Sin2\theta }{g}

R=\frac{381^{2}Sin147 }{9.8}

R = 8067.4 m

Range in shorten by 34.1 %

So, the new range is

R' = 8067.4 - 34.1 x 8067.4/100

R' = 5316.4 m

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Answer:

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Explanation:

First, we find the energy of the electrons in the 8th shell. In order to do this, we recall that the energy of an electron (in the Hydrogen atom) whose principal number is n is given by:

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In order to find the number of subshells we recall that, for a given principal quantum number n, the possible values of the quantum number l, which corresponds to the number of subshells are:

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Since n = 8 in our problem, the possible values of l are: 0, 1, 2, 3, 4, 5, 6, 7. Therefore, the number of subshells are 8.

Now we continue with the number of orbitals. For every subshell l, we have 2l+1 possible values of m, which correspond to the orbitals. Since the possible values of l are: 0,1,2,3,4,5,6,7, therefore, we have to perform the sum:

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Finally, we know that we can fit two electrons per orbital, therefore we can have 64*2 = 128 electrons in the shell corresponding to n=8.

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