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Tcecarenko [31]
3 years ago
14

A particle moving along the x-axis has its position described by the function x=(3.00t3−1.00t 2.00)m where t is in s. at t = 4.0

0 s what is the particle's velocity?
Physics
1 answer:
prohojiy [21]3 years ago
7 0
X =(3.00x4.00 x3-1.00t x 2.00) x m

x= (12.00x3- 1.00 x2.00) x m
x= 36.00 -1.00 x 2.00) x m
x = (36.00 -2.00) x m
x =( 34.00) x m
x =34.00   times m
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Answer: See below

Explanation:

<u>Given:</u>

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a) The normal force is equal to the cos component of the weight of the car.

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2 years ago
Santa doesn't want to push a 25.0 kg wooden box across a wooden floor with a uk of 0.20 at
scZoUnD [109]

Answer:

49 N

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In order to move the box at constant speed, the acceleration of the box must be zero (a=0): this means, according to Newton's second law,

F = ma

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Here there are two forces acting on the box in the horizontal direction while it is moving:

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- The frictional force, F_f

For an object moving on a flat surface, the frictional force is given by

F_f = \mu_k mg

where

\mu_k is the coefficient of friction

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g is the acceleration of gravity

So the equation of the forces becomes

F-\mu_k mg = 0

And substituting:

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