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romanna [79]
4 years ago
13

Two copper wires have the same length but different cross sectional areas. The cross-sectional area of wire 1 is , and that of w

ire 2 is . How are the resistances of the two wires related
Physics
1 answer:
pentagon [3]4 years ago
8 0

Explanation:

We know that resistance

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

clearly,

R∝ 1/A

Greater the area cross section area of the.

therefore,

\frac{R_1}{R_2} = \frac{A_2}{A_1}  =\frac{d_2}{d_1}^2

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Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about the amounts o
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1) At the highest point of the building.

2) The same amount of energy.

3) The kinetic energy is the greatest.

4) Potential energy = 784.8[J]

5) True

Explanation:

Question 1

The moment when it has more potential energy is when the ball is at the highest point in the building, that is when the ball is at a height of 40 meters from the ground. It is taken as a point of reference of potential energy, the level of the soil, at this point of reference the potential energy is zero.

E_{p} = m*g*h\\E_{p} = 2*9.81*40\\E_{p} = 784.8[J]

Question 2)

The potential energy as the ball falls becomes kinetic energy, in order to be able to check this question we can calculate both energies with the input data.

E_{p}=m*g*h\\ E_{p} = 2*9.81*20\\ E_{p} = 392.4[J]\\

And the kinetic energy will be:

E_{k}=0.5*m*v^{2}\\  where:\\v =  velocity = 19.8[m/s]\\E_{k}=0.5*2*(19.8)^{2}\\  E_{k}=392.04[J]

Therefore it is the ball has the same potential energy and kinetic energy as it is half way through its fall.

Question 3)

As the ball drops all potential energy is transformed into kinetic energy, therefore being close to the ground, the ball will have its maximum kinetic energy.

E_{k}=E_{p}=m*g*h = 2*9.81*40\\  E_{k} = 784.8[J]\\ E_{k} = 0.5*2*(28)^{2}\\ E_{k} = 784 [J]

Question 4)

It can be easily calculated using the following equation

E_{p} =m*g*h\\E_{p}=2*9.81*40\\E_{p} =784.8[J]

Question 5)

True

The potential energy at 20[m] is:

E_{p}=2*9.81*20\\ E_{p}= 392.4[J]\\The kinetic energy is:\\E_{k}=0.5*2*(19.8)^{2} \\E_{k}=392[J]

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