Physics, 19.07.2019 02:20 moneykey

What current flows through a 2.56-cm-diameter rod of pure silicon that is 20.0 cm long, when 1.00 ✕ 103 v is applied to it? (such a rod may be used to make nuclear particle detectors, for example.)

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The current flowing through the rod is calculated by the ohm's law which states that the voltage across the circuit is equal to the product of resistance and the current flowing through the circuit.

The current flowing through the rod is $1.11\times 10^{-3}\;\rm A$ .

How do you calculate the current?

Given that the diameter of the rod is 2.56 cm and the length of the rod is 20 cm. The voltage across the circuit is 1.00 ✕ 10^3 V.

We know that the resistivity $\rho$ of silicon is 2300 ohm-m.

The resistance can be calculated as given below.

$R = \rho \dfrac {l}{A}$

Where R is the resistance, l is length and A is the cross-sectional area of the rod.

$R = 2300 \dfrac {0.20}{3.14\times 0.0128^2}$

$R=894145.6\;\rm \Omega$

The current flowing through the rod is calculated as given below.

V = IR

$I = \dfrac {V}{R}$

$I = \dfrac {1\times 10^3}{894145.6}$

$I = 1.11\times 10^{-3}\;\rm A$

Hence we can conclude that the current flowing through the rod is $1.11\times 10^{-3}\;\rm A$ .

To know more about voltage and current, follow the link given below.

link.

Answer from: adantrujillo1234

Current, I = 0.0011 A

Explanation:

It is given that,

Diameter of rod, d = 2.56 cm

Radius of rod, r = 1.28 cm = 0.0128 m

The resistivity of the pure silicon, $\rho=2300\ \Omega-m$

Length of rod, l = 20 cm = 0.2 m

Voltage, $V=1\times 10^3\ V$

The resistivity of the rod is given by :

$R=\rho\dfrac{L}{A}$

$R=2300\ \Omega-m\dfrac{0.2\ m}{\pi (0.0128\ m)^2}$

R = 893692.30 ohms

Current flowing in the rod is calculated using Ohm's law as :

V = I R

$I=\dfrac{V}{R}$

$I=\dfrac{10^3\ V}{893692.30\ \Omega}$

I = 0.0011 A

So, the current flowing in the rod is 0.0011 A. Hence, this is the required solution.

Answer from: Quest

1.the balanced chemical equation is as below

cu + 2 ag(no)₃ → cu(no₃)₂ +2 ag

2.the much silver produced is 50.9 g

step 1: find the moles of cu

moles = mass÷ molar mass

from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236 moles

step 2: use the mole ratio to determine the moles of ag

from equation above cu: ag is 1: 2

therefore the moles of ag =0.236 moles x 2/1 =0.472 moles

step 3: find the mass of ag

mass = moles x molar mass

from periodic table the molar mass of ag =107.87 g/mol

mass = 0.472 moles x 107.87 g/mol = 50.9 g

3. the much silver collected worth is 1.797 oz

convert 50.9 g to oz

that is 1 gram = 0.0353 oz

50.9 grams =? oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1 gram} =1.797 oz

Answer from: Quest

Yup no bull frogs here man

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What current flows through a 2.56-cm-diameter rod of pure silicon that is 20.0 cm long, when 1.00 ✕...

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