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topjm [15]
3 years ago
11

On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total

weight of the balloon, including its load and the hot air in it, is 20,000 N. (a) What is the weight ofthe displaced air? (b) What is the vol- ume ofthe displaced air?
Physics
2 answers:
Tom [10]3 years ago
8 0

Answer:

weight of air displaced by the balloon is 20,000 N

vol- ume ofthe displaced air is  1700 m³

Explanation:

If the balloon is not accelerating, the net force on it is zero.

There are two forces acting on the balloon: the weight (downward)  and the buoyant force of the air (upward).  These must be equal. The buoyant force is then 20,000 N.  The buoyant force equals the weight of air displaced by the balloon, also 20,000 N

The weight of the air displaced is the densty of air times the volume.

The density of air at 1 atm and 20º C is 1.2 kg/m³  

V = 20,000/(1.2*9.8)

=  1700 m³

weight of air displaced by the balloon is 20,000 N

vol- ume ofthe displaced air is  1700 m³

Ilya [14]3 years ago
6 0

Answer:

Explanation:

Since is at test, the buoyancy force, Fb is equal to the weight of the air, W.

W = 20000 N

B.

Density of air = 1.2 kg/m3

Force = mass × Acceleration

= density × volume × acceleration

20000 = 1.2 × Vol × 9.8

Vol = 1700.7 m3.

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Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}.....(I)

Where, F(x) = net force

We know, the net force is the sum of forces.

So, \sum{F}=ma

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

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Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

We need to find the value of \int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

Using newton's second law

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}...(II)

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Acceleration is rate of change of velocity.

a=\dfrac{dv}{dt}

Put the value of a in equation (II)

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\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

Now, the work done by the net force on the block is,

W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

The work done by the net force on the block is equal to the change in kinetic energy of the block.

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Answer:

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We recommend bringing all units to the SI system

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    Iₓ = 17.9 10⁻³ (-0.699 -0)

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    Iₓ = 19.1 10⁻³ (0.745 -0)

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