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AnnyKZ [126]
3 years ago
14

A weather balloon is operating at an altitude at which the density of air is 0.90 kg/m^3. At this altitude, the balloon has a vo

lume of 20m^3 and is filled with helium (density of helium is 0.178 kg/m^3). If the balloon skin weighs 88 N, what load can be supported at this level?
Physics
1 answer:
Archy [21]3 years ago
3 0

Answer:

54 N

Explanation:

Draw a free body diagram.  There are four forces acting on the balloon.  Buoyant force pushing the balloon up, gravity pulling the helium down, gravity pulling the balloon skin down, and gravity pulling the load down.

Apply Newton's second law:

∑F = ma

B − Wh − Wb − L = ma

When the load is at a maximum, the acceleration is 0:

B − Wh − Wb − L = 0

B − Wh − Wb = L

B − mh g − Wb = L

The mass of the helium is its density times its volume:

B − ρh Vh g − Wb = L

Buoyant force is defined as B = ρVg, where ρ is the density of the displaced fluid (in this case, air), V is the volume of the displaced fluid, and g is acceleration of gravity.  Since the volume of displaced air = the volume of the helium:

ρa V g − ρh V g − Wb = L

(ρa − ρh) V g − Wb = L

Given that ρa = 0.90 kg/m³, ρh = 0.178 kg/m³, V = 20 m³, g = 9.8 m/s², and Wb = 88 N:

(0.9 − 0.178) (20) (9.8) − (88) = L

L = 53.5 N

Rounded to 2 sig-figs, the maximum load that can be supported is 54 N.

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B

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A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

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Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

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5 0
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Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r
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Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

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According to ohm's law, V = IR

R = V/I

Here, I = y_{1}

R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms

For the second measurement, current y_{2} = 97 mA = 0.097 A

R = \frac{V}{y_{2} }

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b) y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}

y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]

y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3}  \end{array}\right]

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

x = \left[\begin{array}{ccc}100\end{array}\right]

\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3}  \end{array}\right] =  \left[\begin{array}{ccc}G_{1} \\G_{2}  \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2}  \end{array}\right] =  \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5}  \end{array}\right]

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3 years ago
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