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Rom4ik [11]
3 years ago
8

Can some one please help me with science?

Physics
1 answer:
posledela3 years ago
4 0
Of course I will help you we work as a community on this website
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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
3 years ago
How many molecules of Oxygen gas are there on the reactant side of this equation?
Sergeeva-Olga [200]

Answer:

4

Explanation:

It has 8 O atoms and 4 O2(g) molecules

8 0
2 years ago
A 6- F capacitor is charged to 90 V and is then connected across a 700- resistor. What is the initial charge on the capacitor
Lorico [155]

Answer:

540C.

Explanation:

A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;

Q = CV          ----------(i)

From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.

Where;

C = 6F

V = 90V

Substitute these values into equation (i) as follows;

Q = 6 x 90

Q = 540 C

Therefore, the initial charge on the capacitor is 540C.

7 0
3 years ago
Manipulate p+x=r to solve for x
Tanzania [10]
X=r-p. Maybe I don't understand, but I am assuming that you need to isolate for X? you simply subtract p from both sides.<span />
4 0
3 years ago
Read 2 more answers
On a day when the temperature reaches 50°F, the temperature in degrees Celsius is: 20°C
anzhelika [568]

Answer:

10°C

Explanation:

To convert °F to °C, we use the formula:

°C =  (°F - 32) * ( 5/9)

So, to convert 50°F to the equivalent  in °C, we can proceed as follows:  

°C = ( 50 - 32 ) * (5/9)  

°C = ( 18 ) *  (5/9), which is, approximately,

°C = 9.999999999... ≈ 10 (5/9 ≈0.555555...)

So, 50°F is equivalent to 10°C.

3 0
3 years ago
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