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omeli [17]
3 years ago
15

Which element is found in period 3, group 2?

Physics
2 answers:
kicyunya [14]3 years ago
7 0

Answer:

Mg (Magnesium)

Explanation:

VladimirAG [237]3 years ago
6 0

Answer:  Magnesium (Mg)

Explanation:

I attached a copy of a periodic table for reference.

When we look at the periodic table, go across to group 2 (Alkali Earth Metals) and then down to period 3, where we find Magnesium.

Interestingly, we could also gain this information from Magnesium's spdf electronic configuration: 1s²2s²2p⁶3s². With this, '3' tells us the period, and '²' tells the group.

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A ball of radius R and mass m is magically put inside a thin shell of the same mass and radius 2R. The system is at rest on a ho
dlinn [17]

Answer:

x =\frac{-R}{2}

Explanation:

From the question we are told that mass

Thin layer radius = 2R

Generally the expression for ths solution is given as

Xcm =(m*0 =m(-2R))/2m =-mR/(2m)=-R/2

the center of mass will not move at initial state  

Considering the center of mass of both bodies

xcm=\frac{m*x+m*x)}{2m} =x

x =\frac{-R}{2}

Therefore the enclosing layer moves x =\frac{-R}{2}                          

7 0
3 years ago
The decimal equivalent for meter is
USPshnik [31]

m =dm ______ 10.000

Meters

The metre is a unit of length in the metric system, and is the base unit of length in the International System of Units (SI).

As the base unit of length in the SI and other m.k.s. systems (based around metres, kilograms and seconds) the metres is used to help derive other units of measurement such as the newton, for force.

8 0
3 years ago
A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r
kaheart [24]

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

\Phi = \frac{2\pi \delta}{\lambda}

Where

\delta = Horizontal distance between two points

\lambda = Wavelength

From our values we have,

\lambda = 500nm = 5*10^{-6}m

\theta = 1\°

The horizontal distance between this two points would be given for

\delta = dsin\theta

Therefore using the equation we have

\Phi = \frac{2\pi \delta}{\lambda}

\Phi = \frac{2\pi(dsin\theta)}{\lambda}

\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}

\Phi= 1.096 rad \approx = 1.1 rad

Therefore the correct answer is C.

6 0
3 years ago
A projectile is fired horizontally from a height of 78.4 m at a speed of 300 m/sec. How far did it travel horizontally before hi
Mice21 [21]

Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g

U is the speed = 300m/s

H is the maximum height = 78.4m

g is the acceleration due to gravity = 9.8m/s²

Substitute into the fromula;

R = 300√2(78.4)/9.8

R = 300 √(16)

R = 300*4

R = 1200m

Hence the projectile travelled 1200m before hitting the ground

3 0
3 years ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 47.0 nC placed between q1
lesya [120]

Answer:

Incomplete question, check attachment for completed question

Explanation:

The force of attraction between two forces are given as

F=kQq/r²

4 0
3 years ago
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