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inysia [295]
3 years ago
6

What are two factors affecting friction? How do they affect friction? ch

Physics
1 answer:
AlexFokin [52]3 years ago
5 0
The roughness of the surface, the mass of the object, and the area of contact.
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A cylindrical piece of aluminum is 6.00cm y’all and 2.00cm in radius how much does it weigh
Harlamova29_29 [7]

Answer:

2.00 N

Explanation:

Weight is mass times gravity:

W = mg

Mass is density times volume:

m = ρV

Volume of a cylinder is:

V = πr²h

Finding the volume:

V = π (2.00 cm)² (6.00 cm)

V = 75.4 cm³

The density of aluminum is 2.7 g/cm³.  Finding the mass:

m = (2.7 g/cm³) (75.4 cm³)

m = 204 g

Finding the weight:

W = (0.204 kg) (9.8 m/s²)

W = 2.00 N

The aluminum cylinder weighs 2.00 N.

3 0
2 years ago
I need the letters after the questions
Elodia [21]

Answer:

b d c a

Explanation:

6 0
3 years ago
Monochromatic light is incident on a metal surface, and electrons are ejected. If the intensity of the light increases, what wil
Nadusha1986 [10]
<h2>Answer: The ejection rate will increase.</h2>

The photoelectric effect is a phenomenon that consists of the emission of electrons by certain metals when a beam of light impacts on its surface.

For this phenomenon to occur, certain conditions must be met, such as when the photon collides with the electron, in order to "pull it" from the metal, the photon must have a minimum energy equal to the ionization energy of the atom, so that the electron can leave the influence of the nucleus.  

This is achieved with the adequate intensity of the incident radiation, which is related to the number of photons that impact the metal.

This means:

<h2>The greater the intensity, the greater the number of photons, hence the greater number of electrons emitted.</h2>
6 0
3 years ago
Calculate the force on the test charge by the +6μC charge. Hint: The force in the positive x-direction is positive and the neg
DIA [1.3K]

Answer:

The force on the test charge is 114.75 N.

Explanation:

Given that,

Test charge q=6\ \mu C

Suppose the test charge is 3.4 μC and the distance between the charge and test charge is 4.0 cm.

We need to calculate the force on the test charge

Using formula of electric force

F=\dfrac{kqq_{1}}{r^2}

Where, q = test charge

q₁ =charge

r = distance

Put the value into the formula

F=\dfrac{9\times10^{9}\times6\times10^{-6}\times3.4\times10^{-6}}{(4.0\times10^{-2})^2}

F=114.75\ N

Hence, The force on the test charge is 114.75 N.

6 0
3 years ago
Which unit would be most suitable for its scale?
Cloud [144]

Answer:

Mm, thats the answer trust me men

5 0
3 years ago
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