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4 years ago

14

Grade 11 Physics

1 answer:

xeze [42]4 years ago

6 0

1. Displacement = 50 m North. Distance = 150m
Displacement is a vector. Distance is a scalar.

2. Displacement is 100 miles East.

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For a specific volume of 0.2 m3/kg, find the quality of steam if the absolute pressure is (a) 40 kPa and (b) 630 kPa. What is th

ICE Princess25 [194]

Answer:

$x=0.0498$

$x'=0.659$

Explanation:

Specific Volume $V=0.2m_3/kg$

Absolute Pressure (a) $P_a= 40kpa$

Giving

$T_a=75.87$

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

$v_g=3.993m^3/kg$

(b) $P_a= 630kpa$

Giving

$T_b=160.13C$

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

$v_g'=0.30286 m^3/kg$

(a)

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

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

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

$x=0.0498$

(b)

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

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

$x'=\frac{0.2-1.10*10^{-3}}{3.30-1.1*10^{-3}}$

$x'=0.659$

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3 years ago

This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag

Dmitry [639]

Answer:

Part a)

$F = 135.7 N$

Part b)

$F = 62.5 N$

Explanation:

Part a)

If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block

$Fcos\theta = mg + F_f$

$Fsin\theta = F_n$

so we have

$Fcos\theta = mg + \mu(Fsin\theta)$

$F(cos\theta - \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta - \mu sin\theta}$

$F = \frac{55}{cos50 - 0.310(sin50)}$

$F = 135.7 N$

Part b)

If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block

$Fcos\theta = mg - F_f$

$Fsin\theta = F_n$

so we have

$Fcos\theta = mg - \mu(Fsin\theta)$

$F(cos\theta + \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta + \mu sin\theta}$

$F = \frac{55}{cos50 + 0.310(sin50)}$

$F = 62.5 N$

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