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2 years ago

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

Physics

2 answers:

kotegsom [21]2 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]2 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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Korvikt [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is $\frac{dT}{dt} = 1.016 ^oC/m$

Explanation:

From the question we are told that

The velocity field with which the bird is flying is $\vec V = (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s$

The temperature of the room is $T(x, y, u) = 400 -0.4y -0.6z-0.2(5 - x)^2 \ ^o C$

The time considered is t = 10 \ seconds

The distance that the bird flew is x = 1 m

Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird

Generally the change in the bird temperature with time is mathematically represented as

$\frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 * (5-x)] [-\frac{dx}{dt} ]$

Here the negative sign in $\frac{dx}{dt}$ is because of the negative sign that is attached to x in the equation

$\frac{dT}{dt} = -0.4v_y -0.6v_z -0.2[2 * (5-x)][ -v_x]$

From the given equation of velocity field

$v_x = 0.6x$

$v_y = 0.2t$

$v_z = -1.4$

$\frac{dT}{dt} = -0.4[0.2t] -0.6[-1.4] -0.2[2 * (5-x)][ -[0.6x]]$

substituting the given values of x and t

$\frac{dT}{dt} = -0.4[0.2(10)] -0.6[-1.4] -0.2[2 * (5-1)][ -[0.61]]$

$\frac{dT}{dt} = -0.8 +0.84 + 0.976$

$\frac{dT}{dt} = 1.016 ^oC/m$

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