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4 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]4 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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The force of gravity on a 2-kg rock is twice as great as that on a 1-kg rock. why then doesn't the heavier rock fall faster?

katrin2010 [14]

It takes twice the force to produce the same acceleration in the 2kg rock.

5 0

3 years ago

A well is pumped at Q = 300 m3 /hr in a confined aquifer. The aquifer transmissivity is 25 m2 /hr and the storage coefficient is

Effectus [21]

Answer:

(a). The draw-down at a distance 200 m from the well after pumping for 50 hr is 5.383 m.

(b). The draw-down at a distance 200 m from the well after pumping for 50 hr is 6.707 m.

Explanation:

Given that,

Energy $Q=300\ m^3/hr$

Transmissivity $T = 25\ m^2/hr$

Storage coefficient $S=2.5\times10^{-4}$

Distance r= 200 m

We need to calculate the draw-down at a distance 200 m from the well after pumping for 50 hr

Using formula of draw-down

$s= \dfrac{Q}{4\pi T}(-0.5772-ln\dfrac{r^2S}{4tT})$

Put the value into the formula

$s=\dfrac{300}{4\pi\times25}(-0.5772-ln\dfrac{(200)^2\times2.5\times10^{-4}}{4\times25\times50})$

$s=5.383\ m$

We need to calculate the draw-down at a distance 200 m from the well after pumping for 200 hr

Using formula of draw-down

$s= \dfrac{Q}{4\pi T}(-0.5772-ln\dfrac{r^2S}{4tT})$

Put the value into the formula

$s=\dfrac{300}{4\pi\times25}(-0.5772-ln\dfrac{(200)^2\times2.5\times10^{-4}}{4\times25\times200})$

$s=6.707\ m$

Hence, (a). The draw-down at a distance 200 m from the well after pumping for 50 hr is 5.383 m.

(b). The draw-down at a distance 200 m from the well after pumping for 200 hr is 6.707 m.

3 0

3 years ago

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Artist 52 [7]

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6 0

4 years ago

A 2-kg wood block is pulled by a string across a rough horizontal floor. The string exerts a tension force of 30 N on the block

Luden [163]

Answer:

Work done, W = 84.57 Joules

Explanation:

It is given that,

Mass of the wooden block, m = 2 kg

Tension force acting on the string, F = 30 N

Angle made by the block with the horizontal, $\theta=20^{\circ}$

Distance covered by the block, d = 3 m

Let W is the work done by the tension force. It can be calculated as :

$W=F\ cos\theta\times d$

$W=30\times cos(20)\times 3$

W = 84.57 Joules

So, the work done by the tension force is 84.57 Joules. Hence, this is the required solution.

7 0

3 years ago

C) m1: D) m2: E) r. Please help me I need help now

Nesterboy [21]

A) F = gravitational force

b) G = universal gravitational constant (6.67 × 10-11 N-m2/kg2)

c) m1 = mass of the body 1

d) m2 = mass of body 2

e) r = radius or distance between the two bodies.

Hope this helps!

6 0

3 years ago