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finlep [7]
3 years ago
10

A spherical hot air balloon of diameter 30.0 meters just floats (hovers) when the hot air inside has been heated to a density of

1.10 kg/m3 What is the weight of the balloon and cargo (not including the air inside) if the surrounding air has a density of 1.20 kg/m3? (VSphere = (4/3)nr3)
Physics
1 answer:
Rom4ik [11]3 years ago
5 0

Answer:

W = 166422.729 N

Explanation:

given,

diameter of the balloon = 30 m

density of the air = 1.10 Kg/m³

weight of the balloon and cargo = ?

density of the surrounding air = 1.20 kg/m³

we know,

Density = mass/volume

m = density x volume

m = \rho\times \dfrac{4}{3}\pi r^3

m =1.20 \times \dfrac{4}{3}\pi\times 15^3

m = 16964.6 Kg

Weight of the balloon

 W = m g

 W = 16964.6 x 9.81

W = 166422.729 N

Weight of the balloon and the cargo is equal to  W = 166422.729 N

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4 0
3 years ago
When must scientific theories be changed
madreJ [45]

Answer:

when new information disproving the current theory becomes available.

Explanation:

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2 years ago
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8 0
3 years ago
Read 2 more answers
A box weighing 43.2 N is pulled horizontally until it slides uniformly lat a constant
GREYUIT [131]

Your diagram should include four forces:

• the box's weight, pointing down (magnitude <em>w</em> = 43.2 N)

• the normal force, pointing up (mag. <em>n</em>)

• the applied force, pointing the direction in which the box is sliding (mag. <em>p</em> = 6.30 N, with <em>p</em> for "pull")

• the frictional force, pointing oppoiste the applied force (mag. <em>f</em> )

The box is moving at a constant speed, so it is inequilibrium and the net forces in both the vertical and horizontal directions sum to 0. By Newton's second law, we have

<em>n</em> + (-<em>w</em>) = 0

and

<em>p</em> + (-<em>f</em> ) = 0

So then the forces have magnitudes

<em>w</em> = 43.2 N

<em>n</em> = <em>w</em> = 43.2 N

<em>p</em> = 6.30 N

<em>f</em> = <em>p</em> = 6.30 N

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3 years ago
Under what conditions can an ideal gas undergo a change of state without doing external work?
Cerrena [4.2K]
Nswer 

<span>Work done = Pressure * ΔV [ change in volume ] </span>

<span>If ΔV=0, then no work is done </span>

<span>Zero work thermodyanamic process is called ' isochoric process.' </span>

<span>For example if a gas heated in a rigid container: the pressure and temperature </span>
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<span>This is called an isochoric thermodynamic process. </span>

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