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Alex777 [14]
3 years ago
7

Did I do these questions correctly?

Physics
1 answer:
Elena L [17]3 years ago
8 0
For number 11, you should say that there is more pollution and fossil fuels being burned.
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The dipole moment of HI is 0.42D. What is the dipole moment of HI in C⋅m?
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So in your question where ask to find the dipole moment of HI in C.M. So in my calculation by converting the given data from Debyes to Coulomb Meteres is that 1 Dyebes is equals to 3.33X10^(-30) C.M and the answer would be 1.40X10(-30)C.M. I hope you are satisfied with my answer and feel free to ask for more 
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<h2><u>Projectile</u><u> </u><u>motion</u><u>:</u></h2>

<em>If</em><em> </em><em>an</em><em> </em><em>object is given an initial velocity</em><em> </em><em>in any direction and then allowed</em><em> </em><em>to travel freely under gravity</em><em>, </em><em>it</em><em> </em><em>is</em><em> </em><em>called a projectile motion</em><em>. </em>

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A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig
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Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

V=\frac{MU+mu}{M+m}

Substituting values:

V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134

The coefficient of kinetic friction between the sled and the snow is 0.0134

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