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const2013 [10]
3 years ago
13

A 1.4-kg block slides freely across a rough surface such that the block slows down with an acceleration of â1.25 m/s2. what is t

he coefficient of kinetic friction between the block and the surface?
Physics
1 answer:
Marianna [84]3 years ago
6 0

Mass of the block = 1.4 kg

Weight of the block = mg = 1.4 × 9.8 = 13.72 N

Normal force from the surface (N) = 13.72 N

Acceleration = 1.25 m/s^2

Let the coefficient of kinetic friction be μ

Friction force = μN

F(net) = ma

μmg = ma

μg = a

μ = \frac{a}{g}

μ = \frac{1.25}{9.8}

μ = 0.1275

Hence, the coefficient of kinetic friction is: μ = 0.1275

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A billiards ball B rests on a horizontal surface and is struck by another billiards ball A of the same mass m = 0.2 kg. Ball A i
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Answer:

v1 = 15.90 m/s

v2 = 8.46 m/s

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mechanical energy after collision = 32.433 J

Explanation:

given data

mass m = 0.2 kg

speed = 18 m/s

angle =  28°

to find out

final velocity and  mechanical energy both before and after the collision

solution

we know that conservation of momentum remain same so in x direction

mv = mv1 cosθ + mv2cosθ

put here value

0.2(18) = 0.2 v1 cos(28) + 0.2 v2 cos(90-28)

3.6 =  0.1765 V1 + 0.09389 v2    ................1

and

in y axis

mv = mv1 sinθ - mv2sinθ

0 = 0.2 v1 sin28 - 0.2 v2 sin(90-28)

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from equation 1 and 2

v1 = 15.90 m/s

v2 = 8.46 m/s

so

mechanical energy  before collision = 1/2 mv1² + 1/2 mv2²

mechanical energy before collision = 1/2 (0.2)(18)² + 0

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and

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Answer:

x=0.0498

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Explanation:

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v_f=1.265*10^{-3}m^3/kg

v_g=3.993m^3/kg

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Giving

T_b=160.13C

v_f'=1.10282*10^{-3} m^3/kg

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x=\frac{v-v_f}{v_g-v_f}

x=\frac{0.2-1.0265*10^{-3}}{3.993-1.0265*10^{-3}}

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(b)

Generally the equation for quality of Steam X  is mathematically given by

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