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deff fn [24]
3 years ago
14

An unknown substance is tested and found to have the following physical properties: brittle fair conductor of electricity dull (

not shiny) The substance is most likely a _____. metal metalloid nonmetal cannot be determined without more data
Physics
2 answers:
nikklg [1K]3 years ago
6 0
The answer is <span>metalloid </span>

White raven [17]3 years ago
6 0

Answer:

Metalloid

Explanation:

Metals are good conductors of electricity and are generally shiny.

Metalloids are the elements of the periodic table that have properties of both metals and non metals. They may be shiny or dull. They are brittle and are fair conductors of electricity.

Non metals are dull, do not conduct electricity and brittle.

Hence the substance is most likely a metalloid.

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I need help please so fast
Simora [160]

Answer:

DUumb

Explanation:

Duumb

7 0
3 years ago
The magnitude of the gravitational field strength near Earth's surface is represented by
Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

F = G\cdot \frac{M\cdot m}{r^{2}}

Where:

M - Mass of the planet Earth, measured in kilograms.

m - Mass of the person, measured in kilograms.

r - Radius of the Earth, measured in meters.

G - Gravitational constant, measured in \frac{m^{3}}{kg\cdot s^{2}}.

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

F = m \cdot g

Where:

m - Mass of the person, measured in kilograms.

g - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

g = \frac{G\cdot M}{r^{2}}

Given that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972 \times 10^{24}\,kg and r = 6.371 \times 10^{6}\,m, the magnitude of the gravitational field near Earth's surface is:

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}

g \approx 9.82\,\frac{m}{s^{2}}

The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

4 0
3 years ago
3. Take sugar, oil, corn syrup, a glass and water. Pour the water in the glass and then add each of the above the substances one
hoa [83]

Here are the observations

<u>S</u><u>u</u><u>g</u><u>a</u><u>r</u><u>:</u><u>-</u>

  • Sugar is soluble in water
  • so It will dissolve in water .

<u>C</u><u>o</u><u>r</u><u>n</u><u> </u><u>s</u><u>y</u><u>r</u><u>u</u><u>p</u><u>:</u><u>-</u>

  • Corn syrup is also basically a sugar.
  • It will dissolve in water too .
  • If we shake the mixture in glass then corn syrup will be dissolved.

<u>O</u><u>i</u><u>l</u><u>:</u><u>-</u>

  • Oil is not soluble in water
  • Hence it won't dissolve in water.
  • It will float over water and make two layers
7 0
3 years ago
When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin
mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}

where n=1,2,3 or ...n

L=Length of string

T=Tension

\mu =Mass per unit length

When string is clamped at mid-point

Effecting length becomes L'=0.5 L

Thus new Frequency becomes

f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz      

4 0
3 years ago
Referring to the above diagram, how high will the ball rise on the right-hand incline?
Marina CMI [18]
I think it’s 15cm
Might be 7cm
3 0
4 years ago
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