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marissa [1.9K]
3 years ago
5

What can help protect against the contraction of HIV?

Physics
2 answers:
Gnesinka [82]3 years ago
8 0
Staying protected when you are in contact with other people.. if you understand
Andru [333]3 years ago
6 0
Make sure you have protection and also if you not sure with your partner to get get tested and just ask questions
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Which major planet has the largest . . .
svlad2 [7]

Answer:

Answered

Explanation:

The largest Semi major axis- Neptune( 30.0611 AU).

Largest avg. Orbital speed around the Sun- 47.9 km/s of mercury.

The Largest orbital period around the Sun- 164.79 years for Neptune by 1 year on Earth.

The largest Eccentricity of orbit- is Of Mercury (0.206)

Note: All data has been taken from the internet, hope it helps

7 0
3 years ago
In science, Bob learns that the energy of a wave is directly proportional to the square of the waves amplitude. If the energy of
Naddika [18.5K]

Answer:

Probably none of the above. it's says directly proportional so there would be a proportionality constant.

4 0
3 years ago
The images formed by convex mirrors are always in which form ​
mash [69]

Answer:

Image formed by a <u>convex mirror</u> is always <u>virtual and erect</u>. When an object is placed at infinity, virtual image is formed at focus and the size of the image is <u>smaller</u>.

<h2>도움이되기를 바랍니다!</h2>
4 0
3 years ago
Read 2 more answers
Help I don’t understand this question
7nadin3 [17]
Metals
3/4 of the periodic table
Good conductors of heat and electricity
Malleable

Nonmetals
1/4 of the periodic table
Bad conductors of heat and electricity
Not bendable
5 0
3 years ago
Gauss's law combines the electric field over a surface with the area of the surface. From Coulomb's law we know that the electri
Romashka-Z-Leto [24]

The change in surface area of Gaussian surface with radius (r) is 8πr.

<h3>Electric field from Coulomb's law</h3>

The electric field experienced by a charge is calculated as follows;

E = \frac{Q}{4\pi \varepsilon_o r^2}

where;

  • E is the electric field
  • Q is the charge
  • r is the radius

The electric field reduces by a factor of \frac{1}{r^2}

<h3>Surface area of a Gaussian surface;</h3>

The surface area of a sphere is given as;

A = 4\pi r^2

<h3>Change in area with r</h3>

\frac{dA}{dr} = 8\pi r

Thus, the change in surface area of Gaussian surface with radius (r) is 8πr.

Learn more about area of Gaussian surfaces here: brainly.com/question/17060446

7 0
2 years ago
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