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Angelina_Jolie [31]
2 years ago
12

If one of the satellites is at a distance of 20,000 km from you, what percent accuracy in the distance is required if we desire

a 2-meter uncertainty
Physics
1 answer:
Lesechka [4]2 years ago
8 0
<span>It is quite straightforward to convert an uncertainty to a percent uncertainty. We can divide the amount of uncertainty by the original amount and then multiply by 100%.

(2 m / 20,000,000 m) X 100% = 0.00001%

The percent uncertainty is 0.00001%.

The percent accuracy is the 100% - percent uncertainty.
The percent accuracy = 100% - 0.00001% = 99.99999%

The percent accuracy is 99.99999%.</span>
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Which of the following statements about horizons is true?
nalin [4]
<span>All soils have completely different horizon patterns.</span>
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2 years ago
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Hanging by a thread. Two metal spheres hang from nylon threads and attract each other when brought close together. (i) What can
elena-14-01-66 [18.8K]

Answer:

Explained

Explanation:

i)Two spheres hanging from nylon threads attract each other because either the two spheres are charged with opposite sign or only one of the spheres is charge so the other would be charge by induction of the charged sphere and hence attract each other.

ii)However, when they are touched the charges will be rearranged among the two sphere such that the two sphere have exact same magnitude and sign of charge and now they will repel each other or the  magnitude of charges on the two spheres become zero and they neither attract or repel each other.

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2 years ago
A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi
RSB [31]

Answer:

r = 4.44 m

Explanation:

 

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

         B = ρ g V

Now let's use Newton's equilibrium relationship

         B - W = 0

         B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

         σ = W / A

         W = σ A

The area of ​​a sphere is

           A = 4π r²

       W = W₁ + σ 4π r²

The volume of a sphere is

           V = 4/3 π r³

Let's replace

     ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

     P V = n RT

    P = ρ RT

    ρ = P / RT

 

    P / RT g 4/3 π r³ - σ 4 π r² = W₁

    r² 4π (P/3RT  r - σ) = W₁

Let's replace the values

     r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

     r² (11.81 r -0.060) = 13000 / 4pi

     r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

       r = 4.44 m

3 0
2 years ago
Calculate the flux of the vector field F⃗ =−6i⃗ +5x2j⃗ −5k⃗ , through the square of side 8 in the plane y=1, centered on the y-a
Tasya [4]

Answer:

The flux is 682.6 Wb.

Explanation:

Given that,

Vector field F=-6i+5x^2j-5k

We need to calculate the flux

Using formula of flux

\phi=\int_{-4}^{4}\int_{-4}^{4}(F\cdot j\ dxdz)

Put the value into the formula

\phi=\int_{-4}^{4}\int_{-4}^{4}(-6i+5x^2j-5k)1\ dxdz

\phi=\int_{-4}^{4}\int_{-4}^{4}(5x^2)dxdz

\phi=2(\dfrac{x^3}{3})_{-4}^{4}\times(z)_{-4}^{4}

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7 0
2 years ago
A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d
marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

mgh=\dfrac{1}{2}mv^2

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 35.4}

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0
2 years ago
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