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katen-ka-za [31]

3 years ago

12

A 0.500 m length of wire with a cross-sectional area of 3.14 * 10 ^ - 6 meters squared is found to have a resistance of 2.53 * 1

0 ^ - 3 ohms according to the resisting chart, from what material is the wire made?
PLEASE HELP MEEEEEEE

1 answer:

mr_godi [17]3 years ago

7 0

Answer:

The wire is made of silver (ρ = 1.59×10⁻⁸ ohms/m)

Explanation:

Applying,

R = ρL/A................. Equation 1

Where R = Resistance length of the wire, ρ = Resistivity of the wire, L = Length of the wire, A = crosssectional area of the wire

make ρ the subject of the equation

ρ = RA/L............. Equation 2

From the question,

Given: R = 2.53×10⁻³ ohms, A = 3.14×10⁻⁶ m², L = 0.5 m

Substitute these values into equation 2

ρ = (2.53×10⁻³)(3.14×10⁻⁶)/0.5

ρ = 1.59×10⁻⁸ ohms/m

Hence from the resistivity chart, the wire is made of silver

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Answer:

The charge in each ball will be 3 * 10^-12 C

Explanation:

(Assuming the correct charge of the second ball is 8 * 10^-12)

When the balls are brought in contact, all the charges are split evenly among then.

So first we need to find the total charge combined:

(-3 * 10^-12) + (8 * 10^-12) + (4 * 10^-12) = 9 * 10^-12 C

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To solve this problem we will apply the concepts of equilibrium and Newton's second law.

According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So

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Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

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$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

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Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

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$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

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$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

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$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

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79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

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collect like terms,

2.45366 $T_{f}$ = 283.133

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Learn more about temperature here: brainly.com/question/25677592

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