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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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The 1st Law of Thermodynamics is a statement about (A) Temperatures scales (B) Whether a process can proceed in a specific manne
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Energy conservation.

Explanation:

The 1st Law of Thermodynamics is a statement about energy conservation. It states that \Delta U=Q-W, which means that if we <u>substract the work W done</u> by the system to the <u>heat Q given</u> to the system we get the <u>change in the internal energy</u> \Delta U, so any excess in energy given to the system appears as internal energy, stating that energy is conserved.

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The speed of an arrow fired from a compound
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Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

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Answer:

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5 0
3 years ago
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