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eimsori [14]
3 years ago
14

The force of friction depends upon​

Physics
2 answers:
dezoksy [38]3 years ago
6 0

Answer:

ask internet?

Explanation:

easy .....

..

rosijanka [135]3 years ago
4 0
Friction depends partly on the smoothness of the contacting surfaces, a greater force being needed to move two surfaces past one another if they are rough than if they are smooth.
Therefore, friction depends on the smoothness or roughness of the two surfaces which are in contact.
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What is the relationship between the valence electrons of an atom and the chemical bonds the atom can form?​
stellarik [79]

Answer:

Valence electrons are outer shell electrons with an atom and can participate in the formation of chemical bonds. In single covalent bonds, typically both atoms in the bond contribute one valence electron in order to form a shared pair. The ground state of an atom is the lowest energy state of the atom.

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3 years ago
What is the chemical formula for aluminum nitride?<br> A) Al3 <br> B) AlN3 <br> C) Al3N3 <br> D) AlN
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4 years ago
Read 2 more answers
A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60∘ away from vertical increases fro
bixtya [17]

Answer:

The induced emf is  |\epsilon|=0.0261 V

Explanation:

From the question we are told that

     The number of turn is  N = 100

      The diameter of the coil is  d = 2.0 cm = 2.0 *10^{-2} m

      The uniform magnetic at initial is B_i = 0.50 T

      The uniform magnetic at initial is B_f = 1.50 T

      The time taken is t = 0.60s

      The angle the magnetic field makes with vertical is  \theta = 60^o

Generally induced emf is mathematically represented as

     \epsilon = -N \frac{d \o}{dt}

where d \o is the change  magnetic flux

Magnetic flux is mathematically represented as

       \O = \= B \cdot \= A

          = BA cos \theta

Substituting this above  

       \epsilon = -N \frac{d (BA cos \theta)}{dt}

      \epsilon = -N A \frac{d (B cos \theta)}{dt}

  Where B is the magnetic field and A is the area which is mathematically evaluated as

        A = \frac{\pi  d^2}{4}

Substituting values

        A = \frac{\pi (2.0 *10^{-2})^2}{4}

           A= 3.142*10^{-4}m^2

From the equation of emf

          \epsilon = -N A \frac{d (B cos \theta)}{dt}

dB = B_2 -B_1

       So

             |\epsilon| = N A \frac{ (B_2 -B_1 cos \theta)}{dt}

substituting values

            |\epsilon| = 100(3.142*10^{-4}) \frac{ (1.50 -0.50 cos(60))}{0.60}

                |\epsilon|=0.0261 V

6 0
3 years ago
If you were asked how to evaluate scientific evidence, what would you would want to include in your answer?
raketka [301]
You would want to include qualitative and quantitative evidence
6 0
4 years ago
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 78.5 m/s. Th
Andrews [41]

Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion  (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

-(78.5 m/s)² / 2(250m) = a ⇒ a = -12.3 m/s².

Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

⇒ t= (-78.5 m/s)/ (-12.3 m/s²) ⇒ t= 6.4 seconds.

2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x  (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

3 0
3 years ago
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