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slamgirl [31]
3 years ago
15

Please help me I only have a few questions and need this turned in by the end of the day! In all of our simulations we left the

Coefficient of Friction set to 0. As a result, you should have noticed that the potential energy would transfer directly into kinetic energy as it went down the hill (and vice versa as it went up a hill.) Describe what trends you would expect to see if we included a Coefficient of Friction (in other words, added friction into the simulation).
Physics
2 answers:
Ipatiy [6.2K]3 years ago
6 0

Answer: did you ever get the answer i need help too

Explanation:

jasenka [17]3 years ago
4 0

Answer:

In the absence of non-conservative forces that is friction, the mechanical energy remains conserved. The potential energy changes to kinetic energy as it goes down the hill. At the bottom of the hill

the entire potential energy transforms to kinetic energy. In the presence of dissipative forces like frictional forces, some of the energy is dissipated as heat. So, potential energy does not convert entirely to kinetic energy.

If coefficient of friction was included, the graph would not be straight line but a curved as at the bottom, the kinetic energy is not equal to the potential energy at the top of the hill.

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100 POINTS! I will mark brainliest! Record your hypothesis as an “if, then” statement for the rate of dissolving the compounds:
love history [14]

Answer:

<u><em>Rate of dissolving compounds:</em></u>

If we increase the temperature of the solution, then the dissolving compound would dissolve more easily.

<u><em>Boiling Point of Compounds:</em></u>

If the inter-molecular forces of any compound is really strong, then the boiling point of the compound would be really high.

6 0
4 years ago
An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 34.5
Inessa05 [86]

Answer:

the acceleration of the airplane is 5.06 x 10⁻³ m/s²

Explanation:

Given;

initial velocity of the airplane. u = 34.5 m/s

distance traveled by the airplane, s = 46,100 m

final velocity of the airplane, v = 40.7 m/s

The acceleration of the airplane is calculated from the following kinematic equation;

v² = u² + 2as

2as= v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(40.7)^2 -(34.5)^2}{2 \times 46,100} \\\\a = 5.06 \ \times \ 10^{-3} \ m/s^2

Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²

5 0
3 years ago
Two points charges are brought closer together,increasing the force between them by a factor of 25.By what factor wa their separ
Roman55 [17]

Answer:

The separation between the charges was decreased by a factor of 0.2

Explanation:

The Coulomb's force between two charges is given by;

F = \frac{kq^2}{r^2} \\\\let \ kq^2 \ be \ constant\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\increasing \ the \ force \ between \ them \ by \ factor \ of \ 25\\\\(F_2 = 25F_1)\\\\r_2^2 = \frac{F_1r_1^2}{25F_1}\\\\r_2^2 = \frac{r_1^2}{25}\\\\r_2 = \sqrt{\frac{r_1^2}{25} }\\\\ r_2 = \frac{r_1}{5}

r₂ = 0.2r₁

Therefore, the separation between the charges was decreased by a factor of 0.2.

6 0
3 years ago
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed
kogti [31]

Answer:

a)y_{first}=5.3mm

b)y_{second}=10.6-5.3 =5.3 mm  

Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

y=\frac{m\lambda D}{a} when m is a natural number.

here:

  • m is 1 (to find the central bright fringe)                
  • D is the distance from the slit to the screen
  • a is the slit wide
  • λ is the wavelength

So we have:

y_{first}=\frac{633*10^{9}*3.35}{0.0004}

y_{first}=5.3mm

b)

Now, if we do m=2 we can find the distance to the second minima.

y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}

y_{2}=10.6 mm

Now we need to subtract these distance, to get the width of the first bright fringe :

y_{second}=10.6-5.3 =5.3 mm    

I hope it heps you!

     

4 0
3 years ago
A car accelerates at a constant rate from 0 m/s to 10.0 m/s for 5.00 seconds. What is the displacement?
kirill115 [55]

Answer: A car slows from 22 m/s to 3.0 m/s at a constant rate of 2.1 m/s2. How many seconds are required before the car is a r2.1 m/s2 traveling at 3.0 m/s?

5 pages

Explanation:

8 0
3 years ago
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