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3 years ago

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How does electrical conduction compare between metals, nonmetals, and metalloids

1 answer:

zhuklara [117]3 years ago

4 0

Answer:

metal> metalloids >nonmetals (Electrical conductivity)

Explanation:

Electrical conductivity of objects can be compared by the bonding energy of electrons in them.

Metals have less bonding energy of electrons, so even at room temperature their are significant number of free electrons to carry electrical current.

Nonmetals have a very high bonding energy of electrons, so at room temperature negligible number of free electrons are present so electrical conductivity is very low.

Metalloids have both metallic and non metallic features. The electron bonding energy falls in between that of metals and nonmetals. So electrical conductivity also lies in between metals and nonmetals.

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A bike with 15cm diameter wheels accelerates uniformly from rest to a speed of 7.1m/s over a distance of 35.4m. Determine the an

Finger [1]

Answer:

9.47 rad/s^2

Explanation:

Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,

s = 35.4 m

let a be the linear acceleration.

Use III equation of motion.

v^2 = u^2 + 2 a s

7.1 x 7.1 = 0 + 2 x a x 35.4

a = 0.71 m/s^2

Now the relation between linear acceleration and angular acceleration is

a = r x α

where, α is angular acceleration

α = 0.71 / 0.075 = 9.47 rad/s^2

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3 years ago

MathPhys Help pls Tysm

HACTEHA [7]

Answer:

8.75

Explanation:

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½ mv² = Fd

½ (3.9 kg) (2.9 m/s)² = F (1.4 m)

F = 11.7 N

Next, find the distance at the new velocity.

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½ (3.9 kg) (2.5 × 2.9 m/s)² = (11.7 N) d

d = 8.75 m

3 0

3 years ago

Read 2 more answers

Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of

viktelen [127]

Answer:

$W=2.76\times 10^{-6}\ J$

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

$U=\dfrac{Q^2}{2C}$

Put all the values,

$U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J$

So, the work done is $2.76\times 10^{-6}\ J$ .

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3 years ago

The Moon orbits Earth in an average of p = 27.3 days at an average distance of a =384,000 kilometers. Using Newton’s version of

Korvikt [17]

Answer:

The mass of the earth, $M=6.023\times 10^{24}\ kg$

Explanation:

It is given that,

Time taken by the moon to orbit the earth, $T=27.3\ days=2358720\ m$

Distance between moon and the earth, $r=384000\ km=384\times 10^6\ m$

We need to find the mass of the Earth using Kepler's third law of motion as :

$T^2=\dfrac{4\pi^2}{GM}r^3$

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$M=\dfrac{4\pi^2\times (384\times 10^6)^3}{(2358720)^2\times 6.67\times 10^{-11}}$

$M=6.023\times 10^{24}\ kg$

So, the mass of the earth is $6.023\times 10^{24}\ kg$ . Hence, this is the required solution.

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Yuki888 [10]

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3 years ago