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liberstina [14]
3 years ago
5

A spherical balloon has a radius of 6.35 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 990 kg . Neglect the buoyant force on the cargo volume itself. Required Determine the largest mass of cargo the balloon can lift.
Physics
1 answer:
RSB [31]3 years ago
8 0

Answer: 201 kg.

Explanation:

There are two forces acting (both vertically) on the balloon and the cargo: gravity (downward) and the buoyant force (upward).

If we need that the balloon remain in the air, the buoyant force must be equal to the weight of the balloon and the cargo, as follows:

Fg = Fb

In order to get Fg, we must add the mass of the helium (expressed as a product of the density times the volumen of the balloon, assumed spherical), the mass of the skin and the structure of the balloon, and the mass of the cargo itself, as follows:

Fg = (δHe . 4/3 π r3 + mcargo + mstruct) g (1)

The buoyant force, is equal to the weight of the volumen of the air displaced by the balloon (which is equal to the volumen of the entire balloon as it is completely submerged in air) , which can be written as follows:

Fb = δ air . 4/3 π r3 g (2)

Simplyfing, replacing by the values of δHe, δair, r, and mstruct, and solving for mcargo, we finally get:

mcargo = (4/3 π (6.35m)3 (1.29 kg/m3 – 0.179 kg/m3)) – 990 kg

mcargo =  201 kg.

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On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
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Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

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M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

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The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

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The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

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3 years ago
An acorn falls from a tree and hits the ground in 0.8 s. How far did the acorn fall . Use g = 9.8 m/s^2. Round your final result
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The distance covered by the acorn is 3.136 m.

<u>Explanation:</u>

The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.

Then using the second law of equation,

s=ut+\frac{1}{2}gt^{2}

Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so

   s=0+\left(\frac{1}{2} \times 9.8 \times 0.8 \times 0.8\right)=\frac{6.272}{2}=3.136 \mathrm{m}

So the distance covered by the acorn is 3.136 m.

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