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Dennis_Churaev [7]
2 years ago
8

The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cr

oss-sectional area. According to theory, what should be the ratio of the resistance of the second coil to the first and the fourth coil to the third? Calculate these ratios for your experimental results and compare the agreement with the expected ratio.
Physics
1 answer:
stealth61 [152]2 years ago
6 0

Answer:

\frac{R_2}{R_1}=\frac{A_1}{A_2}\\\frac{R_4}{R_3}=\frac{A_3}{A_4}

Explanation:

The resistance of a conductor is directly proportional to its length and is inversely proportional to its cross-sectional area, this dependence is given by:

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

\rho is the material's resistance, L is the legth and A is the cross-sectional area.

For the first and second coils, we have:

R_1=\frac{\rho L}{A_1}\\R_2=\frac{\rho L}{A_2}\\\rho L=R_1A_1\\\rho L=R_2A_2\\R_1A_1=R_2A_2\\\frac{R_2}{R_1}=\frac{A_1}{A_2}

For the third and fourth coils, we have:

R_3=\frac{\rho L'}{A_3}\\R_4=\frac{\rho L'}{A_4}\\\rho L'=R_3A_3\\\rho L'=R_4A_4\\R_3A_3=R_4A_4\\\frac{R_4}{R_3}=\frac{A_3}{A_4}

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(HELP!!! 30 pts if answered right. )What formula gives the strength of an electric field, E, at a distance from a known source c
umka2103 [35]

Answer:

E=\frac{k\,Q}{d^2}

Explanation:

The strength of an electric field E produced by a single charge Q at a distance d from it is given by the formula: E=\frac{k\,Q}{d^2}, where K represents the Coulomb constant.

Since the electric field E is derived from the Coulomb Force per unit charge using a positive test charge, the field's units will be in units of Newtons/Coulomb, and be the formula for the Coulomb electric force between to charges (Q1 and Q2),

F_C=k\frac{Q_1\,Q_2}{d^2}

but modified with only one charge showing in the numerator of the expression.

8 0
2 years ago
The sun transfers energy to the earth by radiation at a rate of approximately 1.00 kW per square meter of surface.
Mashutka [201]

Answer:

1320336992.2512 m²

1320.33 kilometers or 509.79 miles

Explanation:

Energy transferred by the sun

W=0.24\times 1\times 10^3=240\ W/m^2

Energy required by the United States is 1\times 10^{19}\ J/yr (assumed)

Power

P=\frac{E}{t}\\\Rightarrow P=\frac{1\times 10^{19}}{365.25\times 24\times 3600}\\\Rightarrow P=316880878140.2895\ W

Area

A=\frac{P}{W}\\\Rightarrow A=\frac{316880878140.2895}{240}\\\Rightarrow A=132033699.2512\ m^2

Area of the solar collector would be 1320336992.2512 m²

Converting to km²

1\ m^2=\frac{1}{1000\times 1000}\ km^2

1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1000\times 1000}\ km^2=1320.33\ km^2

Converting to mi²

1\ m^2=\frac{1}{1609.34\times 1609.34}\ mi^2

1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1609.34\times 1609.34}\ mi^2=509.79\ mi^2

Each side of the square would be 1320.33 kilometers or 509.79 miles

4 0
2 years ago
Suppose you first walk 12.0 m in a direction 20 owest of north and then 20.0 m in a direction 40.0osouth of west. How far are yo
alexgriva [62]

Answer:

R=19.5m

\theta = 4.65° S of W

Explanation:

Refer the attached fig.

displacement  of the x and y components

x-component displacement is (R_{x}) = A_{x}+B_{x}

= A \sin(20°) + B \cos(40°)

= -12.0\sin(20°) + 20.0\cos(40°)

= -19.425m

x-component displacement is (R_{y}) = A_{y}+ B_{y}

=  A \cos(20°) - B \sin(40°)

= 12.0\cos(20°) - 20.0\sin(40°)

= -1.579

resultant displacement

∴

R = \sqrt{R_{x}^{2} +R_{y}^{2} }  }

=\sqrt{(-19.425)^{2}+(-1.579)^{2}  }

=19.5m

\theta = \tan^{-1}\left | \frac{R_{x}}{R_{y}} \right |

\theta = \tan^{-1}\left | \frac{1.579}{19.425} \right |

\theta = 4.65° S of W

6 0
2 years ago
The nucleus of an atom consists of protons and neutrons (no electrons). A nucleus of a carbon‑12 isotope contains six protons an
andriy [413]

Answer:

7 electrons

Explanation:

We can solve the problem by using the law of conservation of electric charge: in fact, the total electric charge before and after the collision must be conserved.

Before the collision, we have:

- A nucleus of carbon-12, consisting of 6 protons (charge +1 each) + 6 neutrons (charge 0 each), so total charge of +6

- A nucleus of nitrogen-14, consisting of 7 protons (charge +1 each) + 7 neutrons (charge 0 each), so total charge of +7

So the total charge before the collision is +6+7=+13 (1)

After the collision, we have:

- 17 protons (charge +1 each): total charge of +17

- 4 antiprotons (charge -1 each): total charge of -4

- 7 positrons (charge +1 each): total charge of +7

- 25 neutral particles (charge 0 each): total charge of 0

- N electrons (charge -1 each): total charge of -N

So the total charge after the collision is +17-4+7+0-N=+20-N (2)

Since the charge must be conserved, we have (1) = (2):

+13 = +20 - N

Solving for N,

N = 20 - 13 = 7

So, there are 7 electrons.

4 0
2 years ago
The value of specific heat for copper is 390 J/kg⋅C∘, for aluminun is 900 J/kg⋅C∘, and for water is 4186 J/kg⋅C∘.
abruzzese [7]

Answer:

The equilibrium temperature is

21.97°c

Explanation:

This problem bothers on the heat capacity of materials

Given data

specific heat capacities

copper is Cc =390 J/kg⋅C∘,

aluminun Ca = 900 J/kg⋅C∘,

water Cw = 4186 J/kg⋅C∘.

Mass of substances

Copper Mc = 235g

Aluminum Ma = 135g

Water Mw = 825g

Temperatures

Copper θc = 255°c

Water and aluminum calorimeter θ1= 16°c

Equilibrium temperature θf =?

Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water

McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)

Substituting our data into the expression we have

235*390(255-θf)=

(135*900+825*4186)(θf-16)

91650(255-θf)=(3574950)(θf-16)

23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6

Collecting like terms and rearranging

23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf

8.2*10^6=3.66*10^6θf

θf=80.5*10^6/3.6*10^6

θf =21.97°c

5 0
2 years ago
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