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seropon [69]
3 years ago
14

Which element is located in period 3 of the periodic table?

Physics
1 answer:
liberstina [14]3 years ago
3 0

Answer:

From the given options, the option (D) is correct. Because Al is the aluminium which is located in period 3 of the periodic table.

Explanation:

We know that the 3rd period contains 8 elements which are as follows:

  1. sodium
  2. magnesium
  3. aluminium
  4. silicon
  5. phosphorus
  6. sulfur
  7. chlorine, and
  8. argon

The sodium and magnesium, the first 2 elements, are members of the s-block of the periodic table, while the rest of the members are the elements of the p-block.

Thus, from the given options, the option (D) is correct. Because Al is the aluminium which is located in period 3 of the periodic table.

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Answer:

  • Hence, the work done in moving the charge is 45 J.

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a horizontal force of 300 N is needed to push a crate along the floor at constant speed. What is the friction force on the crate
finlep [7]

Answer:

Since a 100-N force is being used to push the crate in one direction, and there is a total net force of 0, the force of friction acting on the crate must also be 100 N. This frictional force points in the direction opposite of the push.

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A simple harmonic oscillator consists of a block of mass 45 g attached to a spring of spring constant 240 N/m, oscillating on a
Brrunno [24]

Answer:

The maximum displacement ,A= 0.047 m

Explanation:

Given that

Mass, m = 45 g= 0.045 kg

spring constant ,K= 240 N/m

Initial velocity at equilibrium ,v= 3.5 m/s

Lets take the maximum displacement is A

As we know that when the mass reached at the extreme position then the velocity of the mass will become zero.

From energy conservation

\dfrac{1}{2}KA^2=\dfrac{1}{2}mv^2

KA^2=mv^2

A=\sqrt{\dfrac{mv^2}{K}}

Now by putting the values

A=\sqrt{\dfrac{0.045\times 3.5^2}{240}}

A=0.047 m

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Lightbulb A is marked "25 W 120V, "and lightbulb B is marked "100 W 120V " These labels mean that each lightbulb has its respect
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The cost of running the lightbulb A for 30 days at 0.110 per KWh is 1.98

<h3>How to determine the energy </h3>

We'll beging by calculating the energy used by lightbulb A. This can be obtained as follow:

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E = Pt

E = 0.025 × 720

E = 18 KWh

<h3>How to determine the cost for running the bulb for 30 days</h3>

The cost of running the bulb for 30 days can be obtained as follow:

  • Cost per KWh = 0.11
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  • Cost =?

Cost = energy × Cost per KWh

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