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valentina_108 [34]
3 years ago
5

A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a

ir is 1.29 kg/m3. The skin and structure of the balloon has a mass of 950 kg. Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift. Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Aleks [24]3 years ago
3 0

volume of balloon

= 4/3 T R3

= 4/3 x 3.14 x 6.953

= 1405.47 m3

uplift force

= volume of balloon x density of air x 9.8

= = 1405.47 x 1.29 x 9.8

= 1813.05 x 9.8 N

weight of helium gas

= volume of balloon x density of helium x

9.8

= 1405.47 x .179 x 9.8

= 251.58 x 9.8 N

Weight of other mass = 930 x 9.8 N Total weight acting downwards

= 251.58 x 9.8 +930 x 9.8

= 1181.58 x 9.8 N

If W be extra weight the uplift can balance

1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8

1181.58+W=1813.05

W= 631.47 kg

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A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be
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Answer:

p_2 = 664081 N/m^{2}

Explanation:

from the ideal gas law we have

PV = mRT

P = \rho RT

\rho = \frac{P}{RT}

HERE  R is gas constant for dry air  =  287  J K^{-1} kg^{-1}

\rho = \frac{7.00 10^{5}}{287(18+273)}

\rho = 8.38 kg/m^{3}

We know by ideal gas law

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m_1 = \rho V_1 = 8.38 *2*10^{-3}

m_1 = 0.0167 kg

for m_2

m_2 = \rho V_i - V_removed

m_2 = 8.38*(.002 - 10^{-4})

m_2 = 0.0159 kg

WE KNOW

PV = mRT

V, R and T are constant therefore we have

P is directly proportional to mass

\frac{p_2}{p_1}=\frac{m_2}{m_1}

p_2 = p_1 * \frac{m_2}{m_1}

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2 years ago
Need help solving this question.
MatroZZZ [7]

Answer:

See the answers below.

Explanation:

to solve this problem we must make a free body diagram, with the forces acting on the metal rod.

i)

The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.

We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.

For the summation of forces we will take the forces upwards as positive and the negative forces downwards.

ΣF = 0

-15+T-W=0\\T-W=15

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.

ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.

ΣM = 0

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Answer:

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