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sammy [17]
3 years ago
6

A constant force of 85 N accelerates towards a 10 kg box from a speed of 3.0 m/s to a speed of 7.0 m/s as it goes 14 m along a h

orizontal floor. What is the coefficient of friction between the box and the floor?
Mathematics
1 answer:
photoshop1234 [79]3 years ago
5 0

Answer:

Step-by-step explanation:

Let us find the acceleration of box .  

v² = u² + 2as

Putting the values

7² = 3² + 2 a x 14

a = 1.43 m /s²

If coefficient of friction be μ

force of friction = μ mg

= μ x 10 x 9.8

= 98μ

Net force pushing the box

= 85 - 98μ

Applying newton's second law

85 - 98μ = 10 x 1.43

98μ = 85 - 14.3

μ = .72

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Step-by-step explanation:

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7 0
3 years ago
What is the 214 term in the sequence 7,10,13,16? Will thank and rate.
faust18 [17]
 214
*    3
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642
this is your answer but you can tell that 7+3=10 so you just times 3 by 214 that will give you the 214th term in the sequence hope that this helps 

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