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denpristay [2]
3 years ago
8

Elaborate . on what can be learned about the properties by the location of element on the periodic table . Use atomic number 13

as an example .
Physics
2 answers:
abruzzese [7]3 years ago
7 0

Answer:

Explanation:

Location of an element on the periodic table determine the periodic properties of an element such as atomic radius, ionization energy, electron affinity, metallic character, electron negativity,etc.

For Example, Aluminum is the element with atomic number 13 on periodic table. It belong to group 13 i.e. Boron group of the periodic table. Position of aluminum on the periodic table helps in determining it's properties such as, it is soft, silvery-white, non-magnetic and ductile metal. It is a p-block element which lies in period 3 of table with three valence electrons in it's outer most shell.

soldier1979 [14.2K]3 years ago
4 0

Aluminum is a type of metal and that is a properties by the location of element

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9) A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk (M = 0.10 kg, R = 0.10 m) rotating in a horizontal
natima [27]

Answer:

a. ω₂ = 14rad/s

b. ∇K.E = 0.014J

c. The bug does not conserve force while moving on the disk (non-conservative force).

Explanation:

Mass of the bug (m) = 0.02kg

Mass of the cylindrical disk (M) = 0.10kg

Radius of the disk (r) = 0.10m

Initial angular velocity ω₁ = 10rad/s

final angular velocity ω₂ = ?

a.

To calculate the new angular velocity, we relate it to the conservation of angular momentum of the system I.e when the bug was at the edge of the disk and when it is located at the centre of the disk.

I = Mr² / 2

I₁ = Mr₂ / 2 + mr²

I₁ = moment of inertia when the bug was at the edge

I₁ = [(0.10 * 0.10²) / 2 ] + (0.02 * 0.1²)

I₁ = 0.0005 + 0.0002

I₁ = 7.0*10⁻⁴kgm²

I₂ = moment of inertia when yhe bug was at the center of the disk.

I₂ = Mr² / 2

I₂ = (0.01 * 0.01²) 2

I₂ = 0.0005kgm²

for conservation of angular momentum,

I₁ω₁ = I₂ω₂

solve for ω₂

ω₂ = (I₁ * ω₁) / I₂

ω₂ = (7.0*10⁻⁴ * 10) / 5.0*10⁻⁴

ω₂ = 14 rad/s

b. the change in kinetic energy of the system is

∇K = K₂ - K₁

∇K = ½I₂*ω₂² - ½I₁*ω₁²

∇K = ½(I₂*ω₂² - I₁ω₁²)

∇K = ½[(5.0*10⁻⁴ * 14²) - (7.0*10⁻⁴ * 10²)]

∇k = ½(0.098 - 0.07)

∇K = ½ * 0.028

∇K = 0.014J

c.

The cause of the decrease and increase in kinetic energy is because the bug uses a non-conservative force. To conserve the mechanical energy of a system, all the forces acting in it must be conservative.

The work W produced by this force brings the difference in kinetic energy of the system

W = K₂ - K₁

6 0
3 years ago
Can someone help me with this please
andrezito [222]
Carbon: C, 12.011, 6, 12
Oxygen: O, 8, 8, 8, 16
Boron: B, 10.811, 5, 5, 11
3 0
3 years ago
What is the energy of a 4kg apple that is sitting on a 2 m high tree branch?
miv72 [106K]

78.4 joules is the energy of a 4 kg apple that is sitting on a 2 m high tree branch.

<u>Explanation: </u>

When an apple falls to the ground from a tree, its positional energy (stored as potential gravitational energy) turns into kinetic energy, during a fall. Chemical potential energy is chemical energy because it is food and potential energy as it can still have ability to move. So, in the given case, kinetic energy is zero.

To find potential energy, the formula would be

             \text { potential energy }(P . E)=m \times g \times h

Where, given

m – Mass – 4 kg

g-\text { Earth'sgravity }-9.8 \mathrm{m} / \mathrm{s}^{2} (Known value)

h – Height -  2 m

Substitute these values, we get

                  P . E=4 \times 9.8 \times 2=78.4 \text { joules }

8 0
3 years ago
What are the STEPS to using a triple beam balance
Gwar [14]
A good diet. and exersise is the answer
8 0
4 years ago
I'm very confused. Thanks for whoever helps me :)
sergij07 [2.7K]
(C). Remember gravity provides an acceleration of 9.81m/s^2, so the y component of velocity initial is zero because it isn’t already falling, and we have the height, so basically we use the kinematic equation vf^2=vi^2+2ad, substitute given values and you get vf^2=2(9.81)(65) which is 1275, when you take the square root you get 35.7m/s for final velocity
(B). Then you use vf=vi+at to get the equation 35.7=(9.81)t, when you divide out you get 3.64s for time t
(A). Finally, since we assume that there is no acceleration or deceleration horizonatally, we just multiply the time taken for it to hit the ground and the initial speed ((3.64)(35.7)) to get 129.96, with significant figures I would round that to 130 metres.
**this is in the order that I felt was easiest to answer**
6 0
3 years ago
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