What is a core sample?

The voltage between two parallel plates separated by a distance of 3. 0 cm is 120 v. The electric field between the plates is?

Lapatulllka [165]

The electric field between plates is 4000V/m.

An electric field (sometimes E-field) is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles.

The value of the electric field has dimensions of force per unit charge. In the metre-kilogram-second and SI systems, the appropriate units are newtons per coulomb, equivalent to volts per metre.

The voltage between points A and B is

V=E.d

E =V/d (uniform E- field only)

where d is the distance from A to B, or the distance between the plates.

Given:

distance d = 3 cm

voltage V = 120 V

Electric field E = V/d

E = 120 V / 3cm

E = 40 V / 1 cm [ 1 cm = 1/100 m ]

E = 4000 V/m.

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6 0

1 year ago

A rectangular plate, whose streamwise dimension (or chord c) is 0.2 m and whose width (or span b) is 1.8 m, is mounted in a wind

34kurt

Answer:

Answer for the question is given in the attachment.

Explanation:

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4 0

3 years ago

The membrane that surrounds a certain type of living cell has a surface area of 5.1 x 10-9 m2 and a thickness of 1.4 x 10-8 m.

ziro4ka [17]

a) The charge on the outer surface is $1.2\cdot 10^{-12} C$

b) The number of ions is $7.5\cdot 10^6$

Explanation:

a)

The membrane behaves as a parallel plate capacitor, whose capacity is given by the equation

$C=\frac{k\epsilon_0 A}{d}$

where

k = 4.3 is the dielectric constant

$\epsilon_0 =8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$A=5.1\cdot 10^{-9} m^2$ is the surface area

$d=1.4\cdot 10^{-8} m$ is the distance between the two plates

Substituting,

$C=\frac{(4.3)(8.85\cdot 10^{-12})(5.1\cdot 10^{-9})}{1.4\cdot 10^{-8}}=1.4\cdot 10^{-11} F$

The capacity of the membrane is related to the potential difference between the two surfaces by

$C=\frac{Q}{\Delta V}$

where here we have

Q = excess charge on one surface

$\Delta V = 85.5 mV = 0.0855 V$ is the potential difference between the two surfaces

Solving for Q, we find

$Q=C\Delta V=(1.4\cdot 10^{-11})(0.0855)=1.2\cdot 10^{-12} C$

b)

We said that the net charge on the outer surface is

$Q=1.2\cdot 10^{-12} C$

The charge of one K+ ions is equal to the electron charge

$+e=1.6\cdot 10^{-19} C$

Therefore, the number of ions on the outer surface can be found by dividing the total charge by the charge of a single ion:

$N=\frac{Q}{e}=\frac{1.2\cdot 10^{-12}}{1.6\cdot 10^{-19}}=7.5\cdot 10^6$

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6 0

3 years ago

A 3kg book falls from a 2m tall bookshelf what is the speed <br><br> PICTURE included

mafiozo [28]

Answer:

39.2m/s

Explanation:

The potential energy the book has right before it falls is equal to the kinetic energy in falling.

PE = KE

mgh = (1/2)mv

2gh=v

v=(2)(9.81)(2)

v=39.24m/s

8 0

3 years ago

What are the components of the "Earth Radiation Budget?"

Alex777 [14]

Answer:

The energy entering, reflecting, absorbed, and emitted by the earth system are the components of the Earth's radiation budget.

Explanation:

I hope this helps also I hope you have a great day and a new year.

4 0

3 years ago