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3 years ago

Which property do metalloids share with nonmetals

2 answers:

7 0

Metalloids are all solid at room temperature. Some metalloids, such as silicon and germanium, can act as electrical conductors under the right conditions, thus they are called semi-conductors. Silicon for example appears lustrous, but is not malleable or ductile (it is brittle - a characteristic of some nonmetals).

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Anon25 [30]3 years ago

6 0

Answer:

The answer is Metalloids can be opaque as well as nonmetals.

Explanation:

Nonmetals are very different from metals, their surface is opaque, and they are bad conductors of heat and electricity, besides they are low density, and melt at low temperatures. The shape of nonmetals cannot be easily altered, as they tend to be fragile and brittle. On the other hand, metalloids are the group of elements that have the properties of metals and nonmetals, these can be both bright and opaque, and their shape can easily change, the metalloids are conductors of heat and electricity, better way that nonmetals but not as good as metals.

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Two trumpet players are trying to tune their instruments. When they are in tune, they will both be playing the same note and no

marusya05 [52]

Answer:

The second trumpeter will be playing at frequency = 515 Hz

Explanation: Given that the note sounds lower and they can hear 20 beats in 4.0 s.

Beat frequency = 20/4 = 5 Hz

Beat frequency = F2 - F1

5 = 520 - F1

F1 = 520 - 5

F1 = 515 Hz

Since the note sound lower, the second trumpeter will be playing at 515 Hz frequency

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3 years ago

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Answer:

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2 years ago

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Vsevolod [243]

Answer:

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2 years ago

Two point charges are on the y-axis. A 3.0 µC charge is located at y = 1.15 cm, and a -2.28 µC charge is located at y = -2.00 cm

Ghella [55]

Answer:

Total electric potential, $V=1.32\times 10^6\ volts$

Explanation:

It is given that,

First charge, $q_1=3\ \mu C=3\times 10^{-6}\ C$

Second charge, $q_2=-2.28\ \mu C=-2.28\times 10^{-6}\ C$

Distance of first charge from origin, $r_1=1.15\ cm=0.0115\ m$

Distance of second charge from origin, $r_2=2\ cm=0.02\ m$

We need to find the total electric potential at the origin. The electric potential at the origin is given by :

$V=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}$

$V=k(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2})$

$V=9\times 10^9(\dfrac{3\times 10^{-6}}{0.0115}+\dfrac{-2.28\times 10^{-6}}{0.02})$

V = 1321826.08 V

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So, the total electric potential at the origin is $1.32\times 10^6\ volts$ . Hence, this is the required solution.

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2 years ago

How does this relate to Newton's second law?

taurus [48]

Answer:

In ideal case, when no resistive forces are present then both the balls will reach the ground simultaneously. This is because acceleration due to gravity is independent of mass of the falling object. i.e. g = GM/R² where G = 6.67×10²³ Nm²/kg², M = mass of earth and R is radius of earth.

Let us assume that both are metallic balls. In such case, we have to take into account the magnetic field of earth (which will give rise to eddy currents, and these eddy currents will be more, if surface area will be more) and viscous drag of air ( viscous drag is proportional to radius of falling ball), then bigger ball will take slightly more time than the smaller ball.

Explanation:

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3 years ago