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3 years ago

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A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 222 kg and it

is filled with helium gas until its volume is 328 m3. Assume the density of air is 1.20 kg/m3 and the density of helium is 0.179 kg/m3. Calculate the buoyant force acting on the balloon.

1 answer:

inn [45]3 years ago

4 0

Answer:

The buoyant force is 3778.8 N in upward.

Explanation:

Given that,

Mass of balloon = 222 Kg

Volume = 328 m³

Density of air = 1.20 kg/m³

Density of helium = 0.179 kg/m³

We need to calculate the buoyant force acting

Using formula of buoyant force

$F_{b}=\rho_{air}\times V_{b}\times g$

Where, $\rho_{air}$ = density of air

V = Volume of balloon

g = acceleration due to gravity

Put the value into the formula

$F_{b}=1.20\times321\times9.81$

$F_{b}=3778.8\ N$

This buoyant force is in upward direction.

Hence, The buoyant force is 3778.8 N in upward.

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GrogVix [38]

Answer:

x = 0.176 m

Explanation:

For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.

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sin 60 = $T_{y}$ / T

T_{y} = T sin 60

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we apply the equation

∑ τ = 0

-W L / 2 - w x + T_{y} L = 0

the length of the bar is L = 6m

-Mg 6/2 - m g x + T sin 60 6 = 0

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let's calculate

let's use the maximum tension that resists the cable T = 900 N

x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)

x = (4676 - 5880) / 6860

x = - 0.176 m

Therefore the block can be up to 0.176m to keep the system in balance.

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The weight of a luggage is 69.3 N on the moon. Find its weight on the Earth.

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Answer:

Explanation:

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You just calibrated a constant volume gas thermometer. The pressure of the gas inside the thermometer is 294.0 kPa when the ther

Travka [436]

Answer: 361° C

Explanation:

Given

Initial pressure of the gas, P1 = 294 kPa

Final pressure of the gas, P2 = 500 kPa

Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K

Final temperature of the gas, T2 = ?

Let us assume that the gas is an ideal gas, then we use the equation below to solve

T2/T1 = P2/P1

T2 = T1 * (P2/P1)

T2 = (100 + 273) * (500 / 294)

T2 = 373 * (500 / 294)

T2 = 373 * 1.7

T2 = 634 K

T2 = 634 K - 273 K = 361° C

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Which of the following is a device that uses an inclined plane?

Mnenie [13.5K]

Answer:

A and B

Explanation:

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3 years ago

Read 2 more answers

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

so

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago