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3 years ago

12

If the absolute pressure of a gas is 550.280 kPa, its gage pressure is

2 answers:

Rom4ik [11]3 years ago

8 0

The answer is D. 448.955 kPa

aivan3 [116]3 years ago

4 0

By definition we have to:

$Pabs = Patm + Pg$

Where,

Pabs: absolute pressure

Patm: atmospheric pressure

Pg: gage pressure

The atmospheric pressure is constant and its value is:

$Patm = 101.325 kPa$

Then, by clearing gage pressure we have:

$Pg = Pabs - Patm$

Substituting values we have:

$Pg = 550.280 - 101.325\\Pg = 448.955 KPa$

Answer:

If the absolute pressure of a gas is 550.280 kPa, its gage pressure is:

D. 448.955 kPa

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1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

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Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

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a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0

3 years ago

Two force vectors are oriented such that the angle between their directions is 38 degrees and they have the same magnitude. If t

Usimov [2.4K]

Answer:

Sum of the forces will be equal to 3.479 N

Explanation:

We have given two same forces are oriented at an angle of 38°

Magnitude of each force is given $F_1=F_2=1.84$

We have to find the sum of the forces

Sum of the forces will be equal to

$F=\sqrt{F_1^2+F_2^2+2F_1F_2cos\Theta }$

So sum of the forces will be equal to $F=\sqrt{1.84^2+1.84^2+2\times 1.84\times 1.84\times cos38^{\circ} }=3.479$

So sum of the force will be equal to 3.479 N

6 0

3 years ago