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4 years ago

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A 2-m3 rigid tank initially contains air whose density is 1.18 kg/m3 . The tank is connected to a high-pressure supply line thro

ugh a valve. The valve is opened, and air is allowed to enter the tank until the density in the tank rises to 5.30 kg/m3 . Determine the mass of air that has entered the tank?

1 answer:

sdas [7]4 years ago

5 0

Explanation:

It is known that density is the mass present in per unit volume.

Mathematically, Density = $\frac{mass}{volume}$

Since, it is given that $d_{1}$ is 1.18 $kg/m^{3}$ , $d_{2}$ is 5.30 $kg/m^{3}$ , and volume is 2 $m^{3}$ .

Therefore, mass of air that has entered will be $m_{2}$ - $m_{1}$ and it will be calculated as follows.

$d_{2} - d_{1}$ = $\frac{m_{2} - m_{1}}{Volume}$

$m_{2} - m_{1}$ = $(5.30 kg/m^{3} - 1.18 kg/m^{3}) \times 2 m^{3}$

= 8.24 kg

Thus, we can conclude that mass of air that has entered the tank is 8.24 kg.

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In recent years, astronomers have found planets orbiting nearby stars that are quite different from planets in our solar system.

BartSMP [9]

Answer:

The value of g in Kepler-12b is $3.85 m/s^{2}$

Explanation:

To determine the value of g at Kepler-12b it is necessary to combine the equation of the weight and the equation for the Universal law of gravity:

$W = m.g$ (1)

Where m is the mass and g is the value of the gravity

$F = G \frac{M.m}{R^{2}}$ (2)

Equation (1) and equation (2) will be equal, since the weight is a force acting on the object as a consequence of gravity:

$m.g = G \frac{M.m}{R^{2}}$ (3)

Then g will be isolated from equation 3:

$g = G \frac{M.m}{m.R^{2}}$

$g = \frac{G.M}{R^{2}}$ (4)

The radius of Jupiter has a value of 69911000 meters, so its diameter can be determined by:

$d = 2R$ (5)

Where d and R are the diameter and radius of Jupiter

$d_{jupiter} = 2(69911000 m)$

$d_{jupiter} = 139822000 m$

Procedure for finding the radius of Kepler-12b:

For the case of Kepler-12b it has a diameter that is 1.7 times that of Jupiter

$d_{Kepler-12b} = 1.7(d_{jupiter})$

$d_{Kepler-12b} = 1.7(139822000 m)$

$d_{Kepler-12b} = 237697400 m$

By means of equation 5 the radius of Kepler-12b can be known:

$R_{Kepler-12b} = \frac{d}{2}$

$R_{Kepler-12b} = \frac{237697400 m}{2}$

$R_{Kepler-12b} = 118848700 m$

Procedure for finding the mass of Kepler-12b:

The mass of Jupiter has a value of $1.898x10^{27}$ kilograms

For the case of Kepler-12b it has a mass that is 0.43 times that of jupiter

$m_{Kepler-12b} = 0.43(m_{jupiter})$

$m_{Kepler-12b} = 0.43(1.898x10^{27} Kg)$

$m_{Kepler-12b} = 8.161x10^{26} Kg$

The value of g in Kepler-12b can be found by replacing its radius and mass in equation 4:

$g = \frac{(6.67x10^{-11} N.m^{2}/Kg^{2})(8.161x10^{26} Kg)}{(118848700 m)^{2}}$

$g = 3.85 m/s^{2}$

Hence, in Kepler-12b the gravity has a value of $3.85 m/s^{2}$

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3 years ago

A 1.0-in.-diameter hole is drilled on the centerline of a long, flat steel bar that is 1 2 thick and 4 in. wide. The bar is subj

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Answer:

The answers are

The average stress = 20000 lb/in²

The maximum tensile stress immediately adjacent to the hole

= 31076.92 lb/in²

Explanation:

To solve the question we have

Weight of tensile load = 30,000 lb

Width of steel bar = 4 in

Thickness of steel bar = 1/2 in

Average Stress = Force/Area

Size of hole drilled = 1.0 in diameter

Available width at cross section where the 1.00 in diameter hole is drilled =

(4 - 1) in = 3 inches

Cross sectional area at the point of reduced cross section due to the drilled hole = Width × Thickness (Since the item is a flat bar)

= 3 in × 1/2 in = 1.5 in²

Therefore Stress = (30000 lb)/(1.5 in²) = 20000 lb/in²

the maximum tensile stress immediately adjacent to the hole.

Bending stress = $\sigma_B= \frac{M_y}{I}$ where $I = \frac{(0.5^2 + 4^2)}{12}$

0.5*30000/I = 11076.92 lb/in²

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Answer

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$F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}$ when r<R

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