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3 years ago

Pls someone help me pls…. and pls explain to me how

Physics

1 answer:

slavikrds [6]3 years ago

8 0

Answer:

1.12 × 10⁴ m/s

Explanation:

The escape velocity of the object v = √(2GM/R) where G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of the Earth = 6 × 10²⁴ kg and R = radius of the Earth = 6.4 × 10⁶ m

Since v = √(2GM/R)

Substituting the values of the variables into the equation, we have

v = √(2GM/R)

v = √(2 × 6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg/6.4 × 10⁶ m)

v = √(13.34 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg/6.4 × 10⁶ m)

v = √(80.04 × 10⁻¹¹ × 10²⁴Nm²/kg/6.4 × 10⁶ m)

v = √(80.04 × 10¹³Nm²/kg ÷ 6.4 × 10⁶ m)

v = √(80.04 ÷ 6.4 × 10¹³ ÷ 10⁶Nm/kg)

v = √(12.50625 × 10⁷ Nm/kg)

v = √(125.0625 × 10⁶ Nm/kg)

v = 11.18 × 10³ m/s

v = 1.118 × 10 × 10³ m/s

v = 1.118 × 10⁴ m/s

v ≅ 1.12 × 10⁴ m/s

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#from this, the temperature and enthalpy at state 2, $s_2$ can be determined using interpolations $T_2=862K$ and $h_2=890.9kJ/kg$ . The specific volume at $s_1$ can then be determined as:

$\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg$

Specific volume, $s_2$ :

$\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg$

The pressures at $s_2 \ and\ s_3$ is:

$P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa$

.The thermal efficiency=> maximum temperature at $s_3$ can be obtained from the expansion work at constant pressure during $s_2-s_3$

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$a_r_4=\frac{r}{r_c}\alpha _r_3$

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$n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563$

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

$MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa$

Hence, the mean effective relative pressure is 674.95kPa

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3 years ago

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Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

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When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

1 year ago