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Kruka [31]
3 years ago
10

3. Differentiate between: (a) area and volume​

Physics
1 answer:
Margarita [4]3 years ago
8 0

Answer:

area is 2d volume is 3d

Explanation:

Area refers to the size of two-dimensional surface. Volume refers to the size of a three-dimensional space.

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What are the units of radiation?
Ksenya-84 [330]

The radiation dose absorbed by a person is measured using the conventional unit rad or the SI unit gray. The biological risk of exposure to radiation is measured using the conventional unit rem or the SI unit sievert.

I hope this helps!

3 0
3 years ago
Radio waves just like light waves can be reflected refracted and diffracted and polarized.
hoa [83]
<span>Radio waves just like light waves can be reflected refracted and diffracted and polarized.  The answer is True. </span>These characteristics are the common phenomena for electromagnetic (EM)  waves, and Radio Waves are electromagnetic Waves so much so that they obey reflection, refraction, and diffraction. 
5 0
3 years ago
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What is the push and pull of a man about to hit a golf ball with a golf club. Is it causing a change in SPEED, or DIRECTION, or
ivanzaharov [21]
It would be both speed and direction depending on the man's swing 
5 0
4 years ago
Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy
8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

Formula used for the radius of the n^{th} orbit will be,

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV

Energy of the second orbit in H atom .

E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV

Energy of the third orbit in H atom .

E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV

Energy of the fifth orbit in H atom .

E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV

Energy of the sixth orbit in H atom .

E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV

Energy of the seventh orbit in H atom .

E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV

During an absorption of energy electron jumps from lower state to higher state.So,  absorption will take place in :

1) n = 2 → n = 3

2) n=  5 → n = 6

Energy absorbed when: n = 2 → n = 3

E=E_3-E_2

E=(-1.51 eV) -(-3.40 eV)=1.89 eV

Energy absorbed when: n = 5 → n = 6

E'=E_6-E_5

E'=(-0.378 eV)-(-0.544 eV) =0.166 eV

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0
3 years ago
The local high school is installing new bleachers at the stadium and must also add handrails to meet code. The students know the
Advocard [28]

Answer:

The handrails must be approximately 10.63 meters long

Explanation:

The given parameters are;

The height of the bleachers, h = 8 m

The depth of the bleachers, d = 7 m

The length of the hand rails to go along the bleachers from bottom to top is given by Pythagoras' Theorem as follows;

The length of the hand rail = √(d² + h²)

∴ The length of the hand rail = √(7² + 8²) = √113 ≈ 10.63

In order for the handrails to go along the bleachers from top to bottom, they must be approximately 10.63 meters long.

5 0
3 years ago
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