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3 years ago

For metalloids on the periodic table, how do the group number and the period number relate?

Physics

2 answers:

kotegsom [21]3 years ago

8 0

Answer : The correct option is, (1) The lower the group number, the lower the period numbers, so the metalloids are found in a diagonal moving down from left to right.

Explanation :

Metalloids : These are the elements which shows both the property of metals and non-metals that means the metalloids are present in between the metals and non- metals elements.

Boron, silicon, germanium, arsenic, antimony, tellurium and polonium are the metalloids.

In the periodic table, the metals are present on the left side, non-metals are present on the right side and the metalloids are present between the metals and non-metals.

The relation between the group number and the period number is :

The lower the group number, the lower the period numbers. So, the metalloids are found in a diagonal moving down from left to right.

Hence, the correct option is, (1)

The periodic table is also shown below.

lapo4ka [179]3 years ago

3 0

Im guessing it's (a) since the numbers go in chronological order and you read the periodic table left to right

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Answer:

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3 years ago

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3 years ago

Read 2 more answers

The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 5 2 cm, in m/s, is given by u(r) 5 4(1

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The question is not clear and the complete clear question is;

The velocity profile in fully developed laminar flow In a circular pipe of inner radius R = 2 cm, in m/s, is given By u(r) = 4(1 - r²/R²). Determine the average and maximum Velocities in the pipe and the volume flow rate.

Answer:

A) V_max = 4 m/s

B) V_avg = 2 m/s

C) Flow rate = 0.00251 m³/s

Explanation:

A) We are given that;

u(r) = 4(1 - (r²/R²))

To obtain the maximum velocity, let's apply the maximum condition for a single-variable continual real valued problem to obtain;

(d/dr)(u(r)) = 0

Thus,

(d/dr)•4(1 - (r²/R²)) = 0

4(d/dr)(1 - (r²/R²)) = 0

If we differentiate, we have;

4(0 - (2r/R²)) = 0

-8r/R² = 0

Thus, r = 0 and with that, the maximum velocity is at the centre of the pipe.

Thus, for maximum velocity, let's put 0 for r in the U(r) function.

Thus,

V_max = 4(1 - 0²/R²) = 4 - 0 = 4 m/s

B) Average velocity is given by;

V_avg = V_max/2

V_avg = 4/2 = 2 m/s

C) the flow can be calculated from;

Flow rate ΔV = A•V_avg

A is area = πr²

From question, r = 2cm = 0.02m

A = π x 0.02²

Hence,

ΔV = π x 0.02² x 2 = 0.00251 m³/s

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