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1 year ago

12

What does the letter I represent?

1 answer:

irina [24]1 year ago

6 0

The letter i is used to signify that a number is an imaginary number. It stand for the square root of negative one.

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A cup of coffee sits on the dashboard of an automobile. Even though you hold the cup still, as the car goes around a sharp curve

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This would be Newton’s first law :)

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2 years ago

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A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it

Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface $\Phi$ it is possible to calculate the magnitude $\frac{\Phi}{A}=B\ cos \theta$ .

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux $\Phi$ is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is ( $m^{2}$ ). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with $\theta$ being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

$\Phi=B\ A\ cos\theta$

We are told the values of $\Phi$ and B, then we can calculate the magnitude

$\frac{\Phi}{A}=B\ cos\theta$

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2 years ago

How does deformation by drawing of a semicrystalline polymer affect its tensile strength?

Elenna [48]

This effect is explained by increased chain entanglements at higher molecular weights. Increasing the degree of crystallinity of a semicrystalline polymer leads to an enhancement of the tensile strength. Deformation by drawing increases the tensile strength of a semicrystalline polymer.

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2 years ago

A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released

GREYUIT [131]

Answer:

$6.75\mu C/m^2$

Explanation:

We are given that

Diameter,d= $1\mu m=1\time 10^{-6} m$

$1\mu m=10^{-6} m$

Radius,r= $\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m$

Density, $\rho=900kg/m^3$

Total number of electrons,n=39

Charge on electron = $1.6\times 10^{-19} C$

Total charge= $q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C$

Distance,s=2mm= $2\times 10^{-3} m$

Mass = $density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg$

Initial velocity,u=0

Final speed,v=4.5 m/s

$v^2-u^2=2as$

$(4.5)^2-0=2a(2\times 10^{-3})$

$20.25=4a\times 10^{-3}$

$a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2$

Force,F=ma

$qE=ma$

$q(\frac{\sigma}{2\epsilon_0})=ma$

$\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}$

$\epsilon_0=8.85\times 10^{-12}$

$\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2$

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2 years ago

What was significant about the discovery of gallium

Kryger [21]

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