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Lubov Fominskaja [6]
3 years ago
9

20 kg object travels 28 meter and stops. coefficient friction= 0.085 how much work was done by friction?

Physics
1 answer:
Tresset [83]3 years ago
7 0
Assuming it is on a horizontal surface:
friction = μR
R = 20g (g is gravity 9.81)
so Friction = 0.085 x 20g
Work done is force x distance 
so Work done = 0.085 x 20g x 28
 = 466.956 J

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3 years ago
The magnitude of the electrostatic force between two identical ions that are separated by a distance of 5.0A is 3.7×10^-9N.a) wh
jeyben [28]

Explanation:

Given that,

Electrostatic force, F=3.7\times 10^{-9}\ N

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q=\sqrt{\dfrac{Fr^2}{k}}

q=\sqrt{\dfrac{3.7\times 10^{-9}\times (5\times 10^{-10})^2}{9\times 10^9}}      

q=3.2\times 10^{-19}\ C

(b) Let n is the number of electrons are missing from each ion. It can be calculated as :

n=\dfrac{q}{e}

n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}

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Hence, this is the required solution.                        

8 0
3 years ago
The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space b
ZanzabumX [31]

Answer:

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Substitute in these values in the above equation;

V = \frac{2K_e \lambda}{k}ln(\frac{b}{a}) =  \frac{2*8.99*10^9*18*10^6 }{2.3}ln(\frac{3}{0.8}) =140.71 *10^{15} *1.322 \\\\V= 186.02 *10^{15} \ V

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8 0
3 years ago
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UkoKoshka [18]

Answer:

1.1397 Nm

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</span>
6 0
3 years ago
Read 2 more answers
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