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olga2289 [7]
3 years ago
11

Which applied force will allow a 7.65 kg block of ice to begin sliding on a sheet of ice? The block of ice has a kinetic coeffic

ient of friction of 0.030 and a static coefficient of friction of 0.10.

Physics
2 answers:
Debora [2.8K]3 years ago
8 0

Answer:

its D 9.0

Explanation:

i got it wrong so you could get it right ;))

Anni [7]3 years ago
7 0

Answer:

force for start moving is 7.49 N

force for moving constant velocity 2.25 N

Explanation:

given data

mass = 7.65 kg

kinetic coefficient of friction = 0.030

static coefficient of friction = 0.10

solution

we get here first weight of block of ice that is

weight of block of ice = mass  ×  g

weight of block of ice = 7.65 × 9.8 = 74.97 N

so here Ff = Fa

so for force for start moving is

Fa = weight × static coefficient of friction  

Fa = 74.97 × 0.10

Fa =  7.49 N

and

force for moving constant velocity is

Fa =  weight × kinetic coefficient of friction

Fa = 74.97 × 0.030

Fa = 2.25 N

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A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc
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Answer:

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Explanation:

(a)

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(b)

Using,

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Ф = [(0/0.08)+(150/0.08)].100/2

Ф = 93750 rad

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1 rad = 0.159155 rev,

Ф = (93750×0.159155) rev

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