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3 years ago

The elements selenium and oxygen are both poor electrical conductors. In which section of the periodic table are they

Physics

1 answer:

murzikaleks [220]3 years ago

6 0

Answer:

The answer to your question is Nonmetals

Explanation:

Nonmetals they are bad conductors of heat and electricity except graphite.

Metalloids they are less conductors of electricity than metals.

Noble gases they conduct electricity.

Halogens they are not metals and do not conduct electricity.

From this information, we conclude that Oxygen and Selenium are nonmetals.

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Technician A says that torque-to-yield head bolts are special bolts that are reusable. Technician A says that main bearing clear

Blizzard [7]

Answer: D

Neither A nor B

Explanation:

In order to check the clearances for rod and main bearings, you need a set of micrometers and a dial-bore gauge

Measuring the inside diameter of a main or rod bearing will require a dial bore gauge. The best ones to use are accurate down to 0.0001-inch.

So, both technician A and B are incorrect

7 0

3 years ago

1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True B. False

Studentka2010 [4]

1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True

2. The maximum number of electrons allowed in a p sublevel of the 3rd principal level is?
B.6

3. A neutral atom has a ground state electronic configuration of 1s^2 2s^2. Which of the following statements concerning this atom is/are correct?
B. All of the above.

4 0

4 years ago

Read 2 more answers

Biologists use optical tweezers to manipulate micron-sized objects using a beam of light. In this technique, a laser beam is foc

vekshin1

Answer:

Explanation:

Part A) Using

light intensity I= P/A

A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2

Radius= Diameter/2

P= power= 10*10^-3=0.01 W

light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2

Part B) Using

I=c*ε*E^2/2

rearrange to solve for E= $\sqrt{$ ((I*2)/(c*ε))

c is the speed of light which is 3*10^8 m/s^2

ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1

I= the already solved light intensity= 8.85*10^10 W/m^2

amplitude of the electric field E= $\sqrt{$ (9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)

---> E= $\sqrt{$ (1.8*10^11) / (2.66*10^-3) = $\sqrt{$ (6.8*10^13) = 8.25*10^6 V/m

8 0

3 years ago

The capacitor can withstand a peak voltage of 590 volts. If the voltage source operates at the resonance frequency, what maximum

kirill115 [55]

Answer:

The maximum voltage is 41.92 V.

Explanation:

Given that,

Peak voltage = 590 volts

Suppose in an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is $1.20×10^{-2}\ \mu F$ .

We need to calculate the resonance frequency

Using formula of frequency

$f=\dfrac{1}{2\pi\sqrt{LC}}$

Put the value into the formula

$f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}$

$f=2356.88\ Hz$

We need to calculate the maximum current

Using formula of current

$I=\dfrac{V_{c}}{X_{c}}$

$I=2\pi\times f\times C\times V_{c}$

$I=2\pi\times2356.88\times1.20\times10^{-8}\times590$

$I=0.1048\ A$

Impedance of the circuit is

$z=\sqrt{R^2+(X_{L}^2-X_{C}^2)}$

At resonance frequency $X_{L}=X_{C}$

$Z=R$

We need to calculate the maximum voltage

Using ohm's law

$V=I\times R$

$V=0.1048\times400$

$V=41.92\ V$

Hence, The maximum voltage is 41.92 V.

4 0

3 years ago

An electron moving in the direction of the +x-axis enters a magnetic field. If the electron experiences a magnetic deflection in

Gnom [1K]

Answer:

Explanation:

According to the left hand rule for an electron in a magnetic field, hold the thumb of the left hand at a right angle to the rest of the fingers, and the rest of the fingers parallel to one another. If the thumb represents the motion of the electron, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the electron. In this case, the left hand will be held out with the thumb pointing to the right (+x axis), and the palm facing your body (-y axis). The magnetic field indicated by the other fingers will point down in the the -z axis.

5 0

3 years ago