Please need help on this

Can you hellp me please

mamaluj [8]

Answer:

i think the answer is B but im not sure

7 0

3 years ago

Read 2 more answers

If the magnitude of the electric field in air exceeds roughly 3 ✕ 106 N/C, the air breaks down and a spark forms. For a two-disk

Westkost [7]

Answer:

1.843 x 10^-5 C

Explanation:

Givens:

It is given that the air starts ionizing when the electric field in the air exceeds a magnitude of 3 x 10^6 N/C, which means that the max electric field can stand without forming a spark is 3 x 10^6 N/C.

Also it is given that the radius of the disk is 50 cm, it is required to find out the max amount of charge that the disk can hold without forming spark, which means the charge that would produce the max magnitude of the electric field that air can stand without forming spark, and since we know that the electric field in between 2 disk "Capacitor" is given by the following equation

E = (Q/A)/∈o (1)

Where,

Q: total charge on the disk.

A: the area of the disk.

Calculations:

We want to find the quantity of charge on the disk that would produce an electric field of 3 x 10^6 N/C, knowing the radius of the disk we can find the cross-section of the disk, thus substituting in equation (1) we find the maximum quantity of charge the disk can hold

Q = EA∈o

= (3 x 10^6) x (π*0.50) x (8.85 x 10^-12)

= 1.843 x 10^-5 C

note:

calculations maybe wrong but method is correct

8 0

3 years ago

A wire along the z axis carries a current of 6.8 A in the z direction Find the magnitude and direction of the force exerted on a

Alexeev081 [22]

Answer:

Force is 14.93N along positive y axis.

Explanation:

We know that force 'F' on a current carrying conductor placed in a magnetic field of intensity B is given by

$\overrightarrow{F}=\overrightarrow{Il}\times \overrightarrow{B}$

where L is the length of the conductor

Applying values in the equation we have force F =

$\overrightarrow{F}=6.8\times 6.1\widehat{k}\times 0.36\widehat{i}\\\\\overrightarrow{F}=41.48\widehat{k}\times 0.36\widehat{i}\\\\\therefore \overrightarrow{F}=14.93N\widehat{j}$

Thus force is 14.93N along positive y axis.

3 0

3 years ago

A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor

Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

$\Delta A = \frac{1}{2} \Delta \theta R^2$

The formula for the induced emf is

$E = \frac{\Delta \phi}{\Delta t}$

$\phi = \texttt {magnetic flux}$

$E=\frac{\Delta (BA) }{\Delta t}$

$=B\frac{\Delta A}{\Delta t}$

B is the magnetic field strength

substitute

$\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A$

$E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega$

The magnetic field of the earth is oriented at 14.42

$\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5$

we plug in the values in the equation above

so, the induce EMF will be

$E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega$

$=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V$