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victus00 [196]
3 years ago
6

One primary purpose of the European Union is to

Physics
2 answers:
il63 [147K]3 years ago
5 0
The primary purpose of the European Union is to allow more trades between countries. 
I hope it helped :)
Verdich [7]3 years ago
4 0

Answer: The answer is correct (B).

Explanation:

The European Union is an economic bloc aimed at integrating European economies to increase trade between countries and between the bloc and the rest of the world.

Alternatives A, C and D are consequences of the creation of the European Union. The citizens of the bloc have become "European" and therefore can transit and work in all countries of the area, with less bureaucracy. However, the goal behind this is precisely to improve production and business conditions. Likewise, the single currency was created to facilitate this process.

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I am pushing in a large box with 200 N of force. Clint is pushing on the box with 200 N of force in the opposite direction. Why
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Balanced forces

Explanation:

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Why does the frequency of a siren get higher as an ambulance using that siren gets closer?
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61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A
Mademuasel [1]

Answer:

Part a)

percentage = 21.3%

Part b)

percentage = 2.13 \times 10^{-5}%

Explanation:

As we know that total power used in the room is given as

P = P_1 + P_2 + P_3 + P_4

here we have

P_1 = (110)(3) = 330 W

P_2 = 100 W

P_3 = 60 W

P_4 = 3 W

P = 330 + 100 + 60 + 3

P = 493 W

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

110\times i = 493

i = 4.48 A

now resistance of transmission line

R = \frac{\rho L}{A}

R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}

R = 5.23 \ohm

now power loss in line is given as

P = i^2 R

P = (4.48)^2(5.23)

P = 105 W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{105}{493} \times 100

percentage = 21.3%

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

110 \times 10^3 (i ) = 493

i = 4.48\times 10^{-3} A

now power loss in line is given as

P = i^2 R

P = (4.48 \times 10^{-3})^2(5.23)

P = 1.05 \times 10^{-4} W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{1.05 \times 10^{-4}}{493} \times 100

percentage = 2.13 \times 10^{-5}%

6 0
3 years ago
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