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elena55 [62]
3 years ago
5

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. Show that the weight of the displaced air is 20,000 N. b. Show that the volume of the displaced air is 1700 m3 .
Physics
1 answer:
Stella [2.4K]3 years ago
3 0

Explanation:

Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.

Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.

Weight of the air displaced = density of air × volume

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

the volume V = 20,000/(1.2×9.8) =  1700 m³

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Answer: im pretty sure its true

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2 years ago
How much thermal energy is needed to melt 1.25 kg of water at its melting point? Use Q = masslaten heat of fusion.
Amanda [17]

Answer:

Latent heatnof fusion = 417.5 J

Explanation:

Specific latent heat of fusion of water is 334kJ.kg-1.

The heat required to melt water when it's ice I called latent heat because there is no temperature change, the only change observed is change in physical structure.

The amount of heat required to change 1 kg of solid to its liquid state (at its melting point) at atmospheric pressure is called Latent heat of Fusion.

Latent heat = ML

Latent heat= 1.25 kg * 334kJ.kg-1

Latent heat = 1.25*334 *(J/kg)*kg

Latent heat = 417.5 J

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3 years ago
Two long, parallel wires separated by 2.00 cm carry currents in opposite directions. The current in one wire is 1.75 A, and the
Natasha_Volkova [10]

The force per unit length between the two wires is 6.0\cdot 10^{-5} N/m

Explanation:

The magnitude of the force per unit length exerted between two current-carrying wires is given by

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where

\mu_0 = 4\pi \cdot 10^{-7} Tm/A is the vacuum permeability

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r is the separation between the two wires

For the wires in this problem, we have

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I_2 = 3.45 A

r = 2.00 cm = 0.02 m

Substituting into the equation, we find

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m

Learn more about current and magnetic fields:

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4 0
3 years ago
Two cars collide at an intersection. Car A , with a mass of 2000 kg, is going from west to east, while car B, of mass 1500 kg, i
devlian [24]

Answer:

The answer of the part (a) is v2 = 7.09 m/s

and the answer of the part (b) is vA1 = 5.25 m/s

Explanation:

Explanation of the both parts of answer  is in the following attachments

6 0
3 years ago
A worker pushed a 33 kg block 6.1 m along a level floor at constant speed with a force directed 23° below the horizontal. if the
jenyasd209 [6]
The work done occurs only in the direction the block was moved - horizontally. Work is given by:

W = F(h) * d

Where F(h) is the force applied in that direction (horizontal) and d is the distance in that direction. In this case, F(h) is the horizontal component of the applied force, F(app). However, the question doesn't give us F(app), so we need to find it some other way.

Since the block is moving at a constant speed, we know the horizontal forces must be balanced so that the net force is 0. This means that F(h) must be exactly balanced by the friction force, f. We can express F(h) as a function of F(app):

F(h) = F(app)cos(23)

Friction is a little trickier - since the block is being PUSHED into the ground a bit by the vertical component of the applied force, F(v), the normal force, N, is actually a bit more than mg:

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Now we can get down to business and solve for F(app) - as mentioned above:

F(h) = f
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F(app) = 76.8

Now that we have F(app), we can find the exact value of F(h):

F(h) = F(app)cos(23)
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And now that we have F(h), we can find W:
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W = 431.3

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3 0
3 years ago
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