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3 years ago

If Bob applies 10 N of force on box, what will happen to the acceleration of the box if he adds more weight to it?

Physics

2 answers:

monitta3 years ago

4 0

The acceleration will decrease

forsale [732]3 years ago

3 0

A. is your answer hope this helps.

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Consider a Hydrogen atom with the electron in the n 8 shell. What is the energy of this system? (The magnitude of the ground sta

Shtirlitz [24]

Answer:

The energy of an electron in the 8th shell is given by: -0.2125 eV

The number of subshells is: 8

The number of orbitals is: 64

The number of electrons that fit on this shell is: 128

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:

$E_{n}=-13.6\frac{1}{n^{2}}$

Substituting n=8, we find that the energy is given by:

$E_{8} = -13.6\frac{1}{8^{2}}=-0.2125$

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:

0, 1, 2, ... , n-1

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:

$\sum_{l=0}^{7}(2l+1) = 8^2=64$

And we can conclude that the number of orbitals is equal to 64.

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.

8 0

4 years ago

A current I flows down a wire of radius a.

Helga [31]

Answer:

(a) $K = \frac{I}{2\pi a}$

(b) $J = \frac{I}{2\pi as}$

Explanation:

(a) The surface current density of a conductor is the current flowing per unit length of the conductor.

$K = \frac{dI}{dL}$

Considering a wire, the current is uniformly distributed over the circumferenece of the wire.

$dL = 2\pi r$

The radius of the wire = a

$dL = 2\pi a$

The surface current density $K = \frac{I}{2\pi a}$

(b) The current density is inversely proportional

$J \alpha s^{-1}$

$J = \frac{k}{s}$ ......(1)

k is the constant of proportionality

$I = \int\limits {J} \, dS$

$I = J \int\limits \, dS$ ........(2)

substituting (1) into (2)

$I = \frac{k}{s} \int\limits\, dS$

$I = k \int\limits^a_0 \frac{1}{s} {s} \, dS$

$I = 2\pi k\int\limits\, dS$

$I = 2\pi ka$

$k = \frac{I}{2\pi a}$

substitute $J = \frac{k}{s}$

$J = \frac{I}{2\pi as}$

7 0

3 years ago

Pls help on this one

lidiya [134]

Answer:

Explanation:

because 50.0/10.0 = 5.0

5 0

3 years ago

Read 2 more answers

The law of reflection says that the angle of incidence is

nekit [7.7K]

The law of reflection states that the angle of incidence is equal to the angle of reflection. Furthermore, the law of reflection states that the incident ray, the reflected ray and the normal all lie in the same plane.

hope this helps :)

4 0

3 years ago

An asteroid with a mass of 5.0 x 105 kg collides with the Earth and slides horizontally along the ground without bouncing (not a

V125BC [204]

The strength of the friction doesn't matter. Neither does the distance or the time the asteroid takes to stop. All that matters is that the asteroid has

1/2 (mass) (speed squared)

of kinetic energy when it lands, and zero when it stops.
So

1/2 (mass) (original speed squared)

is the energy it loses to friction in order to come to rest.

8 0

3 years ago