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Ksju
1 month ago
11

1. A U.S. manufacturer wants to sell computer equipment to companies in Honduras. Because the U.S. government limits trade with

Honduras, the manufacturer is required to obtain an export license from the Department of Commerce before selling and shipping the equipment to the purchasers. The export license fee is $8,100, not including a nonrefundable application fee of $2,700. What are the possible effects of these fees?
Physics
1 answer:
Ostrovityanka [3.2K]1 month ago
3 0
If the U.S. government prohibits the manufacturer from exporting their products to Honduras, the manufacturer will forfeit the non-refundable application fee of $2,700. Additionally, due to the restrictions on trade with Honduras, the manufacturer must secure an export license from the Department of Commerce, potentially leading to reduced demand for U.S. goods. Furthermore, the total value of product exports must reach at least $10,800 while the license is active to justify the application and licensing fees.
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A small mass m is tied to a string of length L and is whirled in vertical circular motion. The speed of the mass v is such that
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(mv^2/R)/(mg)=1/2

v^2=R/2g

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No official standard exists for extinguisher-cylinder colors. that said, a red cylinder is typically associated with which type
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A skateboarder is attempting to make a circular arc of radius r = 16 m in a parking lot. The total mass of the skateboard and sk
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To address this question, we will utilize concepts linked to centripetal force, aligning it with the static frictional force acting on the object. Using this relationship, we can derive the velocity and input the known values. The defined values are:

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What is an example of a renewable resource?
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Read 2 more answers
Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller
Softa [3030]

Answer:

σ₁ = 3.167 * 10^{-6} C/m²

σ₂ = 7.6 * 10 ^{-6}  C/m²

Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface  ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )

358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }

Charge (Q₁) = 572.8 * 10^{-9} C

Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.

V = constant

∴\frac{Q_{1} }{R} = \frac{Q_{2} }{r}

Q_{2} = \frac{r}{R} *Q_{1}

Q_{2} = \frac{5}{12} *572.8*10^{-9}   = 238.666 *10^{-9} C

Surface charge density ( σ₁ ) for the larger sphere.

Calculated Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = \frac{Q_{1} }{A_{1} } = \frac{572.8 *10^{-9} }{0.180864} = 3.167 * 10^{-6}  C/m².

Surface charge density ( σ₂ ) for the smaller sphere.

Calculated Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  = 0.0314 m².

σ₂ =\frac{Q_{2} }{A_{2} } = \frac{238.66 *10^{-9} }{0.0314} = 7.6 * 10 ^{-6} C/m²

8 0
2 months ago
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