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3 years ago

Name three categories that are used to classify the elements in the periodic table?

Physics

2 answers:

dusya [7]3 years ago

7 0

Metals, non-metal and noble gases are the three categories useful for classifying the elements in periodic table.

<u>Explanation:</u>

Elements are defined as a vertical group and a horizontal period in periodic table. Metals, non-metal and noble gases are the three categories useful for classifying the elements in periodic table.

A large number of elements in periodic table flow under the Metals category. Non-metal in the table are mostly gaseous. Noble gases in the table are colorless and non-reactive.

nikitadnepr [17]3 years ago

6 0

Answer:

metals,nonmetals, and inert gases

Explanation:

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Three identical capacitors are connected in parallel to a battery. if a total charge of q flows from the battery, how much charg

muminat

Each capacitor carry the same charge 'q'.

Discussion:
The voltage from the battery is distributed equally across all of the capacitors when they are linked in series. The three identical capacitors' combined voltage is computed as follows:

$V_{T}$ = V₁ +V₂ +V₃

This voltage may also be calculated using capacitance and charge;

V = Q/ C

$V_{T}$ = V₁ +V₂ +V₃

Provided that the total charge is 'q', hence the total voltage can be expressed as:

$V_{T}$ = (Q/C₁) + (Q/C₂) + (Q/C₃) = Q(1/C₁ +1/C₂ +1/C₃)

Therefore from the above explanation, it is concluded that each and every capacitor carry same charge 'q'.

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1 year ago

A finch rides on the back of a Galapagos tortoise, which walks at the stately pace of 0.060 m/s. After 1.1 minutes, the finch ti

Romashka [77]

Answer:

Average Speed = 6.37 m/s

Explanation:

The average speed is simply given by the following formula:

Average Speed = Total Distance Traveled/Total Time Spent

here,

Total Time Spent = 1.1 min + 1.5 min = (2.6 min)(60 s/min) = 156 s

Now, for total distance, we have to calculate the distance traveled on tortoise and distance traveled while flying, separately. Therefore,

Distance Traveled on Tortoise = (Time spent on Tortoise)(Speed of Tortoise)

Distance Traveled on Tortoise = (1.1 min)(60 s/min)(0.06 m/s) = 3.96 m

Similarly,

Flying Distance = (Flying Time)(Flying Speed) = (1.5 min)(60 s/min)(11 m/s)

Flying Distance = 990 m

Since, total distance is the sum of both distances, therefore,

Total Distance = 3.96 m + 990 m = 993.96 m

Now, using the values in equation of average speed, we get:

Average Speed = 993.96 m/156 s

<u>Average Speed = 6.37 m/s</u>

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3 years ago

The beginning development of a star is marked by a supernova explosion, with the gases present in the nebula being forced

GenaCL600 [577]

The beginning development of a star is marked by a supernova explosion, with the gases present in the nebula being forced to scatter. As the star shrinks, radiation of the surface increases and create pressure on the outside shell to push it away and forming a planetary nebula or white dwarf.

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3 years ago

A bullet leaves a gun with a horizontal velocity of 100 m/s. Calculate the distance it would travel in 1.3 seconds.

stiks02 [169]

Ignoring air resistance, the bullet's horizontal velocity is constant:

$v_x=v_{0x}=100\,\dfrac{\mathrm m}{\mathrm s}$

In 1.3 seconds, we can expect it to travel

$v_xt=\left(100\,\dfrac{\mathrm m}{\mathrm s}\right)(1.3\,\mathrm s)=130\,\mathrm m$

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3 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

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3 years ago