1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Aleonysh [2.5K]
2 years ago
5

Can someone help me please

Physics
1 answer:
WINSTONCH [101]2 years ago
8 0

Metalloids

Metalloids have properties that are intermediate between metals and non-metals. They have properties that are difficult to characterize.

Physical Properties of Metalloids

State of matter    -     Solid

Lustre              -      Metallic lustre

Elasticity              -      Brittle

Conductivity      -      Semi-conductive in nature

Chemical Properties of Metalloids

Oxidation             -      They are capable of forming glasses.

Alloys                   -       When mixed with metals, they form alloys.

Allotropic             -        Metallic and non-metallic allotropes are formed.

Melting                 -        Few metalloids contract when they are melted.

Compounds         -   Compounds are formed when they react with halogens.

Common Properties of Metalloids

  • Metalloids are good semiconductors.
  • The reactivity of metalloids is dependent on the properties of the elements they are reacting with.
  • Electronegativity and ionization energy are between metals and non-metals.

Non-Metals

Non-metals are those which lack all the metallic attributes. They are good insulators of heat and electricity. They are mostly gases and sometimes liquid. Some they are even solid at room temperatures like Carbon, sulphur and phosphorus.

Properties of Non metals

Characteristic properties of non-metals are high ionization energies and high electronegativity. Owing to these properties, non-metals usually gain electrons when reacting with other compounds, forming covalent bonds. Among the non-metals, the anionic dopants have a strong influence on the VB. Non-Metal dopants are carbon, nitrogen, fluorine, sulphur and iodine.

General properties of non-metals.

  • The atoms of non-metals tend to be smaller than those of metals. Several of the other properties of non-metals result from their atomic sizes.
  • Non-metals exhibit very low electrical conductivities. The low or non-existent electrical conductivity is the most important property that distinguishes non-metals from metals.
  • Non-metals have high electronegativities. This means that the atoms of non-metals have a strong tendency to attract more electrons than what they would normally have.
  • Non-metals have high electronegativities. This means that the atoms of non-metals have a strong tendency to hold on to the electrons that already have. In contrast, metals rather easily give up one or more electrons to non-metals, metal therefore easily form positively charged ions, and metals readily conduct electricity.
  • Under normal conditions of temperature and pressure, some non-metals are found as gases, some found as solids and one is found as liquid. In contrast, except mercury, all metals are solids at room temperature. The fact that so many non-metals exist as liquids or gases means that non-metals generally have relatively low melting and boiling points under normal atmospheric conditions.
  • In their solid-state, non-metals tend to be brittle. Therefore, they lack the malleability and ductility exhibited by metals.

Physical Properties of Non-Metals

  • Ductility is the property of the material to be stretched into wires but non-metals are not ductile except for carbon, as carbon fibres find uses in a wide variety of industries including sports and music equipment.
  • Another property characteristic to metals is absent in non-metals called malleability. They can’t be drawn into sheets as they are brittle and break on applying pressure.
  • They are not lustrous as they do not have any shiny appearance.
  • They are not sonorous and do not produce a deep ringing sound when they are hit with another material. They are also bad conductors of heat and electricity except for graphite.

You might be interested in
A ball is thrown vertically upwards with a velocity of 30m/s. Determine the maximum height reached
padilas [110]
The maximum height reached is 45.92 m

6 0
2 years ago
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po
gregori [183]

Answer:

E = 0    r <R₁

Explanation:

If we use Gauss's law

      Ф = ∫ E. dA = q_{int} / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

           E = 0    r <R₁

6 0
2 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
When the current in a toroidal solenoid is changing at a rate of 0.0240 A/s , the magnitude of the induced emf is 12.4 mV . When
Gemiola [76]

Answer:

The number of turns in the solenoid is 230.

Explanation:

Given that,

Rate of change of current, \dfrac{dI}{dt}=0.0240\ A/s

Induced emf, \epsilon=12.4\ mV=12.4\times 10^{-3}\ V

Current, I = 1.5 A

Magnetic flux, \phi=0.00338\ Wb

The induced emf through the solenoid is given by :

\epsilon=L\dfrac{dI}{dt}

or

L=\dfrac{\epsilon}{(di/dt)}........(1)

The self inductance of the solenoid is given by :

L=\dfrac{N\phi}{I}.........(2)

From equation (1) and (2) we get :

\dfrac{\epsilon}{(di/dt)}=\dfrac{N\phi}{I}

N is the number of turns in the solenoid

N=\dfrac{\epsilon I}{\phi (dI/dt)}

N=\dfrac{12.4\times 10^{-3}\times 1.5}{0.00338 \times 0.024}

N = 229.28 turns

or

N = 230 turns

So, the number of turns in the solenoid is 230. Hence, this is the required solution.

3 0
2 years ago
A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the ri
lawyer [7]

Answer:

x = 0.6034 m

Explanation:

Given

L = 5 m

Wplank = 225 N

Wman = 522 N

d = 1.1 m

x = ?

We have to take sum of torques about the right support point.  If the board is just about to tip, the normal force from the left support will be going to zero.  So the only torques come from the weight of the plank and the weight of the man.

∑τ = 0  ⇒     τ₁ + τ₂ = 0  

Torque come from the weight of the plank = τ₁

Torque come from the weight of the man = τ₂

⇒  τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)

⇒  τ₂ = Wman*x = 522 N*x   (clockwise)

then

τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0

⇒  x = 0.6034 m

7 0
2 years ago
Other questions:
  • A car owner forgets to turn off the headlights of his car while it is parked in his garage. If the 12.0-V battery in his car is
    10·2 answers
  • An open vertical tube has water in it. a tuning fork vibrates over its mouth. as the water level is lowered in the tube, a reson
    14·1 answer
  • Data that shows consistent, regular, repetitive form displays a pattern.<br><br> True<br><br> False
    12·2 answers
  • A wire 6.90 m long with diameter of 2.15 mm has a resistance of 0.0320 Ω. Find the resistivity of the material of the wire. rho
    7·1 answer
  • A 306 g cart moves on a horizontal, frictionless surface with a constant speed of 14.2 cm/s. A 76.3 g piece of modeling clay is
    7·1 answer
  • A ball with an initial velocity of 25 m/s is subject to an acceleration of -9.8m/s^2 how high does it go before coming to a mome
    8·1 answer
  • If an amplitude modulation (AM) has an amplitude modulating signal of 0.5 V and carrier amplitude of 1.3 V, what is the modulati
    7·1 answer
  • An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 61.0 cm mark on the track. The glider compl
    6·1 answer
  • The product of voltage times amperage is what
    14·2 answers
  • If we interpret the large, angular rocks to have originated from the outcrop at the top of the hill, we are using __________ rea
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!