All categories

3 years ago

The pattern of repeating properties of elements revealed in the periodic table is known as the (Blank)

2 answers:

4 0

<span>The pattern of repeating properties of elements revealed in the periodic table is known as the periodic law.</span>

Paraphin [41]3 years ago

3 0

It is known as periodic law

You might be interested in

If a Styrofoam ball and a steel ball are the exact same size & shape,which has more mass?

Zinaida [17]

Answer:

Steel

Explanation:

Mass is density times volume.

m = ρV

Since they have the same size and shape, they have the same volume.

Steel has a higher density than Styrofoam, so at the same volume, the steel ball will have more mass.

7 0

3 years ago

Read 2 more answers

Two examples of a stopping motion

Anton [14]

Pause?Freeze?
Stop?
Halt?

5 0

2 years ago

Read 2 more answers

in a young's double-slit experiment the center of a bright fringe occurs wherever waves from the slits differ in the distance th

blagie [28]

Answer:

Zero

Explanation:

Because using

Deta X= dsinစ x n(lambda)

But we know that for central maxima

n is zero

So after substituting

Deta x = 0

3 0

3 years ago

brick, dry soil, paper, and water. If all four substances were exposed to sunlight for the same amount of time, which substance

denis-greek [22]

Water because the light is able to enter the water and allow heat to enter faster.

4 0

3 years ago

A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotating t

koban [17]

Answer:

The coefficient of static friction between the object and the disk is 0.087.

Explanation:

According to the statement, the object on the disk experiments a centrifugal force due to static friction. From 2nd Newton's Law, we can represent the object by the following formula:

$\Sigma F_{r} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}$ (1)

$\Sigma F_{y} = N - m\cdot g = 0$ (2)

Where:

$N$ - Normal force from the ground on the object, measured in newtons.

$m$ - Mass of the object, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v$ - Linear speed of rotation of the disk, measured in meters per second.

$R$ - Distance of the object from the center of the disk, measured in meters.

By applying (2) on (1), we obtain the following formula:

$\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}$

$\mu_{s} = \frac{v^{2}}{g\cdot R}$

If we know that $v = 0.8\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $R = 0.75\,m$ , then the coefficient of static friction between the object and the disk is:

$\mu_{s} = \frac{\left(0.8\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.75\,m)}$

$\mu_{s} = 0.087$

The coefficient of static friction between the object and the disk is 0.087.

5 0

3 years ago