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Mice21 [21]
3 years ago
9

What’s the difference between the resistance of 1,000 feet of aluminum No. 14 wire and 1,000 feet of No. 14 copper wire? A. The

aluminum has 0.84 ohms more resistance. B. The aluminum has 2.66 ohms more resistance. C. The copper has 0.59 ohms more resistance. D. The aluminum has 1.99 ohms more resistance.
Physics
1 answer:
Lapatulllka [165]3 years ago
3 0

the answer is A. The aluminum has 0.84 ohms more resistance.

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Multiple-Concept Example 6 reveiws the principles that play a role in this problem. A nuclear power reactor generates 2.3 x 109
r-ruslan [8.4K]

Answer:

change in mass = 2.41*10^{8}kg

Explanation:

The change in the mass can be computed by using the relation

E=\Delta mc^2\\\Delta m=\frac{E}{c^2}(1)

That is, the energy liberated comes from the mass of the nuclear fuel. The energy generated in one year is

E=Pt=2.3*10^{9}\frac{J}{s}*1 year*\frac{365.25 day}{1 year}*\frac{24h}{1 day}*\frac{3600s}{1h}=7.25*10^{16}J

Hence, by replacing in the equation (1) you have  (c=3*10^{8}m/s)

\Delta m=\frac{7.25*10^{16}J}{3*10^{8}\frac{m}{s}}=2.41*10^{8}kg

HOPE THIS HELPS!!

3 0
3 years ago
Read 2 more answers
Need help finding the average speed.
Snowcat [4.5K]

Explanation:

To find the average of these numbers, we just have to add the three numbers together and divide by 3.

  • 2.07 + 0. 74 + 1.33 = 4.14. 4.14 / 3 = 1.38
  • 1.09 + 1.40 + 0.31 = 2.8. 2.8 / 3 ≈ 9.3333333/ 9 1/3
  • 0.95 + 1.61 + 0.56 = 3.12 / 3 = 1.04
  • 0.81 + 1.89 + 1.08 = 3.78 / 3 = 1.26
5 0
3 years ago
Help me find the acceleration
ANEK [815]

a = 3.09 m/s²

<h3>Explanation</h3>

This question doesn't tell anything about how long it took for the car to go through 105 meters. As a result, the <em>timeless </em>suvat equation is likely what you need for this question.

In the <em>timeless</em> suvat equation,

a = \dfrac{v^2 - u^2}{2\; x}

where

  • a is the acceleration of the car;
  • v is the <em>final</em> velocity of the car;
  • u is the <em>initial</em> velocity of the car; and
  • x is the displacement of the car.

Note that <em>v</em> and <em>u</em> are velocities. Make sure that you include their signs in the calculation.

In this question,

  • a is the unknown;
  • v = -10.9 \; \text{m} \cdot \text{s}^{-2};
  • u = -27.7 \; \text{m} \cdot \text{s}^{-2}; and
  • x = - 105 \; \text{m}.

Apply the <em>timeless</em> suvat equation:

a = \dfrac{v^{2} - u^{2}}{2\; x}\\\phantom{a} = \dfrac{(-10.9)^{2} - (-27.7)^{2}}{2 \times (-105)}\\\phantom{a} = 3.09 \; \text{m} \cdot \text{s}^{-2}.

The value of a is greater than zero, which is reasonable. Velocity of the car is negative, meaning that the car is moving backward. The car now moves to the back at a slower speed. Effectively it accelerates to the front. Its acceleration shall thus be positive.

7 0
3 years ago
A beam of electrons is accelerated from rest through a potential difference of 0.400 kV and then passes through a thin slit. Whe
Oksi-84 [34.3K]

Answer:

v= 103.5 V; energy =1.65 x 10^-17

Explanation:

the deflected energy eV sin θ

8 0
2 years ago
Read 2 more answers
Suppose you have three identical metal spheres, AA, BB, and CC. Initially sphere AA carries a charge qq and the others are uncha
Sedaia [141]

Complete Question

Suppose you have three identical metal spheres, A, B, and C. Initially sphere A carries a charge q and the others are uncharged. Sphere A is brought in contact with sphere B, and then the two are separated. Spheres CC and BB are then brought in contact and separated. Finally spheres AA and CC are brought in contact and then separated. What is the final charge on the sphere B, in terms of q?

a. 3/8q

b. 1/4q

c. 3/4q

d. q

e. 5/8q

f. 1/3q

g.1/2q

h. 0

Answer:

   The correct option is b

Explanation:

From the question we are told that

          The charge carried by A is  q C

           The charge carried by B is 0 C

            The charge carried by C is 0 C

When A and B are brought close and then separated the charge carried by  A and B is mathematically evaluated as

                 \frac{ 0 + q}{2} =   \frac{q}{2}

When C and B are brought close and then separated the charge carried by  C and B  is mathematically evaluated as    

                    \frac{0 + \frac{q}{2} }{2}  = \frac{q}{4}

When C and A are brought close and then separated the charge carried by  C and A  is mathematically evaluated as  

                       \frac{\frac{q}{4} +  \frac{q}{2} }{2}   = \frac{3q}{8}

Looking at these calculation we can see that the charge carried by B is

        \frac{q}{4} C

8 0
2 years ago
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