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2 years ago

5

What current flows through a 2.54cm diameter rod of pure silicon that is 20cm long when 1000V is applied?

1 answer:

vfiekz [6]2 years ago

3 0

Answer: $0.0039\ A$

Explanation:

Given

Diameter of the rod $d=2.54\ cm$

length of rod is $l=20\ cm$

Resistivity of silicon is $\rho=6.4\times 10^2\ \Omega-m$

cross-section of the rod $A$

$\Rightarrow A=\dfrac{\pi d^2}{4}\\\\\Rightarrow A=\dfrac{3.142\times 2.54^2\times 10^{-4}}{4}\\\\\Rightarrow A=5.067\times 10^{-4}\ m^2$

Resistance of rod is R

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

$\Rightarrow R=\dfrac{640\times 0.20}{5.067\times 10^{-4}}\\\\\Rightarrow R=25.26\times 10^4\ \Omega$

Current is given by

$\Rightarrow I=\dfrac{V}{R}\\\\\Rightarrow I=\dfrac{1000}{25.26\times 10^4}\\\\\Rightarrow I=0.0039\ A$

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A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot

denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

$P_r = 345 Watt$

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

$P_r = 345 Watt$

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

$\Delta V = E + i r$

here we have

$E = 12 V$

$i = 56 A$

$r = 0.11$

now we have

$\Delta V = 12 + (0.11)(56) = 18.16 V$

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

Part c)

Rate of energy conversion into EMF is given as

$P_{emf} = i E$

$P_{emf} = (56)(12)$

$P_{emf} = 672 Watt$

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

$V = E - i r$

$V = 12 - (56)(0.11)$

$V = 12 - 6.16 = 5.84 V$

Part e)

now the rate of energy dissipation is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

7 0

3 years ago

Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How fa

zloy xaker [14]

Answer:

150000000000 m

0.0000005 seconds

33.33 ns

Explanation:

Speed of electromagnetic waves through vacuum = $3\times 10^8\ m/s$

Echo time = 1000 seconds

Echo time is the time taken to reach the object and come back to the observer

Distance = Speed×Time

$Distance=3\times 10^8\times \frac{1000}{2}=150000000000\ m$

Venus is 150000000000 m away from Earth

Time = Distance / Speed

$Time=\frac{75}{3\times 10^8}=0.00000025\ seconds$

Echo time will be twice the time

$Echo\ time=2\times 0.00000025=0.0000005\ seconds$

The echo time will be 0.0000005 seconds

Difference in time = Difference in distance / Speed

$\Delta t=\frac{\Delta d}{v}\\\Rightarrow \Delta t=\frac{10}{3\times 10^8}\\\Rightarrow \Delta=33.33\ ns$

The accuracy by which I will be able to measure the echo time is 33.33 ns

5 0

3 years ago

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance

jeka94

Answer:

$1.8\times 105 N/C$

Explanation:

We are given that

$u=2\times 10^7 m/s$

$v=4\times 10^7 m/s$

d=1.9 cm= $\frac{1.9}{100}=0.019 m$

Using 1m=100 cm

We have to find the electric field strength.

$v^2-u^2=2as$

Using the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

$16\times 10^{14}-4\times 10^{14}=0.038a$

$0.038a=12\times 10^{14}$

$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Mass of electron,m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Substitute the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

7 0

3 years ago

Can someone pls help me

natali 33 [55]

Answer:

if not mistaken its b

5 0

2 years ago

Read 2 more answers

A stiff wire 32.5 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the l

Flauer [41]

Answer: i personally think it would be 789

Explanation:

Hopefully I am right

4 0

3 years ago