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zhuklara [117]
3 years ago
13

Definition of main energy level

Physics
2 answers:
svlad2 [7]3 years ago
8 0

Answer:

the orbital in which the electron is located relative to the atom's nucleus.

Explanation:

Aleks [24]3 years ago
5 0

Answer:

the principal energy level of an electron refers to the shell or orbital in which the electron is located relative to the atom's nucleus.

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Use heavy-duty grounded extension cords. Heavy-duty cords will have a marking on the insulation such as ___.
Shalnov [3]

Heavy-duty cords will have a marking on the insulation such as letters which indicates its properties.

<h3>What are Heavy duty cords?</h3>

Heavy duty cords comprises of insulators which helps prevent cases of electric shock. There are usually letters written on them which indicates the nature of insulation and other properties.

For example, the  "S" in "SO" stands for "Extra Hard Service and the "O" stands for oil resistant.

Read more about Cords here brainly.com/question/13600515

7 0
3 years ago
The tendency for an object to sink or float has to do with the object's density.<br> True<br> False
Stella [2.4K]

Answer:

TRUE

Explanation:

BECAUSE AN OBJECTS DENSITY  SHOWS HOW MUCH WEIGHT IS PER CUBIC UNIT. LIKE SAD HAS MORE DENSITY THAN WATER SO IT SINKS WHILE WOOD HAS LESS DENSITY THAN WATER SO IT FLOATS

4 0
3 years ago
3. If I run 150m in 30 seconds, what speed will I have been running at?
Radda [10]

Answer:

speed = distance/time

Explanation:

speed = 150/30

speed =5m/s

you were running fast .....5m/s is a good speed

7 0
3 years ago
Lead hs a density of 11.3 g/cm3 and a mass of 282.5 g . what is its volume?
JulsSmile [24]
D = m / V

11.3 = 282.5 / V

V = 282.5 / 11.3

V = 25 cm³
4 0
4 years ago
If a certain brand of solar panels is rated at a value of 1.50 KW/m2 , and a person needed to generate 2.50 MJ in an hour, what
alisha [4.7K]

Answer:

A=0.462\ m^2

Explanation:

Power rating of a solar panel is 1.50 KW/m²

It generates 2.50 MJ in an hour.

We need to find the area of this type of solar panel would be needed. The power pertaining to generate this energy is given by :

P=\dfrac{2.5\times 10^6}{1\ h}\\\\P=\dfrac{2.5\times 10^6}{3600\ s}\\\\P=694.44\ W

Let A be the area of the solar panel. It is calculated as follows :

\dfrac{P}{A}=1.5\times 10^3\\\\A=\dfrac{694.44}{1.5\times 10^3}\\A=0.462\ m^2

So, the required area of the solar panel is 0.462\ m^2.

4 0
3 years ago
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