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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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Point charges q1 and q2 are separated by a distance of 60 cm along a horizontal axis.

amm1812

Answer:

38 cm from q1(right)

Explanation:

Given, q1 = 3q2 , r = 60cm = 0.6 m

Let that point be situated at a distance of 'x' m from q1.

Electric field must be same from both sides to be in equilibrium(where EF is 0).

=> k q1/x² = k q2/(0.6 - x)²

=> q1(0.6 - x)² = q2(x)²

=> 3q2(0.6 - x)² = q2(x)²

=> 3(0.6 - x)² = x²

=> √3(0.6 - x) = ± x

=> 0.6√3 = x(1 + √3)

=> 1.03/2.73 = x

≈ 0.38 m = 38 cm = x

8 0

3 years ago

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snow_lady [41]

Answer:

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Explanation:

6 0

3 years ago

Can you explain hooks law in simple words?

dedylja [7]

Answer:

The extension of a material or a spring is its increase in length when pulled. Hooke’s Law says that the extension of an elastic object is directly proportional to the force applied to it. In other words:

Explanation:

7 0

2 years ago

Read 2 more answers

A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,

alexgriva [62]

The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
$S= \frac{1}{2} a t^2$
where a is the acceleration. Using the data of the problem, we can find a:
$a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2$
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J$

and so the power output is
$P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W$

b) power output at t=4.0s
The velocity at t=4.0 s is
$v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J$

and so the power output is
$P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W$

c) Power output at t=6.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J$

and so the power output is
$P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W$

8 0

3 years ago

It takes Desmond from 4:00 to 9:00 to swim 27 km. What is his speed?

Hatshy [7]

The formula for speed is distance÷time. The distance in this situation is 27km and the amount of time taken to cover that distance is 5 hours. Now, we substitute the values into an equation:
Speed = 27km/5h
Speed = 5.4km/h
Therefore, Desmond's speed is 5.4km/h.

3 0

3 years ago