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3 years ago

Why do we say that Neptune was the first planet to be discovered through the use of mathematics?

Physics

1 answer:

puteri [66]3 years ago

7 0

Answer:

explained

Explanation:

the Neptune was the first planet discovered through the use of mathematics by two astronomers one French and other English. This was breakthrough success in the field of astronomy that marked the importance of mathematics in astronomy. The discovery of the Neptune resulted from the need to explain the motion of Uranus, motion of which could not be explained by the gravitational effect of Jupiter and Neptune.It needed very complex mathematical equations to be Solved to explain it. The two astronomers were Joseph le Verrier and John Couch Adams.

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A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0

2 years ago

The same force is applied to two skateboards. One rolls across the room and the other moves a few feet and comes to a stop. Wher

igor_vitrenko [27]

The longer you spend reading and thinking about this question,
the more defective it appears.

-- In each case, the amount of work done is determined by the strength
of the force AND by the distance the skateboard rolls while you're still 
applying the force. Without some more or different information, the total
distance the skateboard rolls may or may not tell how much work was done
to it.

-- We know that the forces are equal, but we don't know anything about
how far each one rolled while the force continued. All we know is that
one force must have been removed.

-- If one skateboard moves a few feet and comes to a stop, then you
must have stopped pushing it at some time before it stopped, otherwise
it would have kept going.

-- How far did that one roll while you were still pushing it ?

-- Did you also stop pushing the other skateboard at some point, or
did you stick with that one?

-- Did each skateboard both roll the same distance while you continued pushing it ?

I don't think we know enough about the experimental set-up and methods
to decide which skateboard had more work done to it.

8 0

2 years ago

Read 2 more answers

In one of the original Doppler experiments, a tuba was played at a frequency of 64.0 Hz on a moving flat train car, and a second

wolverine [178]

Answer:

$f_{beat} = 1.64\ Hz$