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Juliette [100K]
3 years ago
5

The two hot-air balloons in the drawing are 48.2m and 61.0 m above the ground.A person in the left balloon observes that the rig

ht balloon is 13.3 degrees above the horizontal.What is the horizontal distance x between the two balloons?

Physics
1 answer:
mafiozo [28]3 years ago
4 0

Answer:

The horizontal distance x between the two balloons is 54.15 m

Explanation:

The diagram described as obtained online is presented in the image attached to this solution.

Let the horizontal distance between the two balloons be x

Difference in height (vertical distance) between the two balloons = 61 - 48.2 = 12.8 m

Using trigonometric relations, it is evident that

Tan 13.3° = 12.8/x

x = 12.8/tan 13.3° = 12.8/0.2364 = 54.15 m

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An apple falling to the ground is not an example of centripetal acceleration.
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A proton with charge 1.601019 As is moving at 2.4105 m/s through a magnetic field of 4.5 T. You want to find the force on the pr
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The force on the proton is 17.4 N.

<h3>What is the force on the proton?</h3>

Now we know that the proton is positively charged and that the force on the charge as it moved through the magnetic field could be given by the relation; F = qvB

Where;

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Having said this, we can see that;

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Read 2 more answers
The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
Nikolay [14]

Answer:

Index of expansion: 4.93

Δu = -340.8 kJ/kg

q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

The ideal gas equation:

p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

20 C = 293 K

p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

The variation of specific internal energy is:

Δu = Cv * Δt

Δu = 0.71 * (20 - 500) = -340.8 kJ/kg

The first law of thermodynamics

q = l + Δu

The work will be the expansion work

l = p2*v2 - p1*v1

35 bar = 3500000 Pa

7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

7 0
3 years ago
Three balls, with masses of 3m,2m and m, are fastened to a massless rod of length L. The rotational inertias about the ledt
Oduvanchick [21]

I = MR^2

The Attempt at a Solution:::

I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2

I total = 3ML^2/2

It says the answer is 3ML^2/4 though.

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mark it as brainliest.... ✌✌✌

8 0
3 years ago
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