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3 years ago

What element atomic number is in Group 17 and Period 6?

Physics

1 answer:

notsponge [240]3 years ago

7 0

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The halogen elements are the six elements in Group 17 of the periodic table. Group 17 is the second column from the right in the periodic table and contains six elements: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (As), and tennessine (Ts).

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All matter is made up of subatomic particles that contain atoms.TrueFalse

antiseptic1488 [7]

All matter is made up of subatomic particles that contain atoms. This is false statement.

Everything in the universe is made of matter, and so, everything in the universe is made of atoms. An atom itself is made up of three tiny kinds of particles called subatomic particles: <u>protons, neutrons, and electrons</u>.

All matter is made up of substances called elements, which have specific chemical and physical properties and cannot be broken down into other substances through ordinary chemical reactions.

Matter is anything which have mass and occupy some volume.

On earth, solid, liquid, and gas are the most common states of matter. Not only is water the most common substance on earth, but it is also the only substance that commonly appears as a solid, a liquid, and a gas within the normal range of earth's temperatures.

Learn more about matter here:- brainly.com/question/16982523

#SPJ9

3 0

1 year ago

A penny has a mass of 2.50 g, a diameter of 19.55 mm, and a thickness of 1.55 mm. Calculate the density of the material from whi

Dmitry [639]

Density = (mass) divided by (volume)

We know the mass (2.5 g). We need to find the volume.

The penny is a very short cylinder.
The volume of a cylinder is (π · radius² · height).
The penny's radius is 1/2 of its diameter = 9.775 mm.
The 'height' of the cylinder is the penny's thickness = 1.55 mm.

Volume = (π) (9.775 mm)² (1.55 mm)

   = (π) (95.55 mm²) (1.55 mm)

   = (π) (148.1 mm³)

   = 465.3 mm³

We know the volume now. So we could state the density of the penny,
but nobody will understand what we have. Here it is:

   mass/volume = 2.5 g / 465.3 mm³ = 0.0054 g/mm³ .

Nobody every talks about density in units of ' gram/(millimeter)³ ' .
It's always ' gram / (centimeter)³ '.
So we have to convert our number for the volume.

   (0.0054 g/mm³) x (10 mm / cm)³

   = (0.0054 x 1,000) g/cm³

   = 5.37 g/cm³ .

This isn't actually very close to what the US mint says for the density
of a penny, but it's in a much better ball park than 0.0054 was.

4 0

3 years ago

A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal

zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0

3 years ago

A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten

babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

$P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2$

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0

4 years ago

Thiết bị trao đổi nhiệt ngược chiều, nhiệt độ chất lỏng nóng vào là 310°C, ra là 115°c. Nhiệt độ chất lỏng lạnh vào là 25°c và r

Vika [28.1K]

Answer: 129.5 m

Explanation:

310 + 115 + 25 + 68 = 518

518 / 4 = 129.5 m

i think. Sorry if this is wrong

5 0

3 years ago