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4 years ago

A block of mass 27.00 kg sits on a horizontal surface with, coefficient of kinetic

Physics

1 answer:

zhannawk [14.2K]4 years ago

5 0

Answer:

The force is $F = 172 \ N$

Explanation:

From the question we are told that

The mass of the block is $m_b = 27.0 \ kg$

The coefficient of static friction is $\mu_s = 0.65$

The coefficient of kinetic friction is $\mu_k = 0.50$

The normal force acting on the block is

$N = m * g$

substituting values

$N = 27 * 9.8$

$N = 294.6 \ N$

Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as

$F_f = \mu_s * N$

substituting values

$F_f = 0.65 * 264.6$

$F_f = 172 \ N$

Now for this block to move the force require is equal to $F_f$ i.e

$F= F_f$

=> $F = 172 \ N$

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4 years ago

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natka813 [3]

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3 years ago

A single loop of area 0.193 m2 is placed perpendicular to a magnetic field of 0.374 T. The loop is rotated about an axis lying i

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Given that;

Area A = 0.193 m2

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Using the formula

E.m.f = NABwcosØ

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Therefore, the angular speed w of the loop = 181 rad/s

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3 years ago

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3 years ago