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dybincka [34]
4 years ago
12

A block of mass 5 kg is placed on a rough table having a coefficient of static friction μs = 0.2.What is the force required to m

ove the block?
Physics
1 answer:
just olya [345]4 years ago
5 0

Answer:

<em>We need to (at least) apply a force of 9.8 N to move the block</em>

Explanation:

<u>Second Newton's Law</u>

If a net force F_n different from zero is applied to an object of mass m, then it will move at an acceleration a, given by

F_n=ma

If we apply a force F to an object placed on a rough surface, the only way to make it move is to beat the friction force  which is given by

F_r=\mu F_N

Where \mu is the static friction coefficient and F_N is the normal force exerted by the table to the object. Since there is no motion in the vertical direction the normal force equals the weight of the object:

F_N=mg=5\ kg\ 9.8\ m/s^2=49\ N

The friction force is

F_r=0.2 (49)=9.8\ N

Thus, we need to (at least) apply a force of 9.8 N to move the block

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93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures
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Answer:

828 kg/m³ or 0.828 g/cm³

Explanation:

Applying,

D = m/V............. Equation 1

Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.

From the question,

Given: m = 77 g , V = 93 cm³

Substitute these values into equation 1

D = 77/93

D = 0.828 g/cm³

Converting to kg/m³

D = 828 kg/m³

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