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2 years ago

Much power does a 10 Ohm bulb have in a series circuit with a battery of 12V and nothing else in the the circuit?

Physics

1 answer:

Sedaia [141]2 years ago

8 0

Answer:

it will be d) 14.4W

Explanation:

potential difference (v) = 12 volts

resistance (r) = 10 ohms

now, we know

=》

$power = \frac{v {}^{2} }{r}$

=》

$power \: = \frac{12 {}^{2} }{10}$

=》

$power = \frac{144}{10}$

=》

$power = 14.4 \: watt$

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An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

3 years ago

If the magnitude of the magnetic field is 6.50 mT at a distance of 12.8 cm from a long straight current carrying wire, what is t

Lunna [17]

Answer: magnitude of the magnetic field at a distance of 19.4 cm from the wire=4.29mT

Explanation:

According to Biot-Savart law, A magnetic field generated by a current carrying wire at a distance is represented as

B=μ₀I/ 2πr

B = magnetic field intensity 1000 mT =1T, 6.50mT = 6.50 X 10^-3T

μ₀ =permeability of free space 4π × 10−7 H/m

I = current intensity

r = radius, 100cm = 1m, 12.8 cm= 12.8 x 10^-2m

6.50 X 10^-3 = μ₀ x I/ 2 π X 12.8 X 10^-2

I =6.50 X 10 ^-3 X 2π X X 12.8 X 10^-2/ 4π × 10−7 H/m

I= 4160 A

when the magnetic field is at 19.4 cm from the wire

B=μ₀I/ 2πr

= 4π × 10−7 H/m x4160/ 2π x 19.4 x 10^-2

=0.004288

= 4.29x 10 ^-3T

= 4.29mT

7 0

3 years ago

The resistivity of gold is at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 m

IgorC [24]

Answer:

0.0360531138247 V/m

Explanation:

$\rho$ = Resistivity of gold = $2.44\times 10^{-8}\ \Omega .m$ (General value)

I = Current = 940 mA

d = Diameter = 0.9 mm

A = Area = $\dfrac{\pi}{4}d^2$

E = Electric field

Resistivity is given by

$\rho=\dfrac{EA}{I}\\\Rightarrow E=\dfrac{\rho I}{A}\\\Rightarrow E=\dfrac{2.44\times 10^{-8}\times 940\times 10^{-3}}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}\\\Rightarrow E=0.0360531138247\ V/m$

The electric field in the wire is 0.0360531138247 V/m

6 0

3 years ago

Read 2 more answers

Calculate, using a diagram, the resultant of the following vector combinations:

Veronika [31]

Add them together with south being negative. (-350 + 25) to get 325 south

4 0

3 years ago

7. These temperatures were recorded in Pasadena for a week in April. 87 85 80 78 83 86 90 Find each of these. (a) Mean (e) Range

slega [8]

Answer:

a) Mean = 84.14

b) Median = 85

c) Mode = no mode (since there is no variable that appears more than once in this dataset)

d) Midrange = 84

e) Range = 12

f) Variance = 14.69

g) Standard deviation = 3.83

Explanation:

The raw data to be processed is

87 85 80 78 83 86 90

a) Mean = (Σx)/N

The mean is the sum of variables divided by the number of variables

x = each variable

N = number of variables = 7

Mean = (87+85+80+78+83+86+90)/7

Mean = 84.14

b) Median is the number in the middle of the dataset when the variables are arranged in ascending or descending order.

Arranging the data in ascending order

78, 80, 83, 85, 86, 87, 90

The number in the middle is the 4th number = 85

Median = 85

c) Mode is the variable that occurs the most in a distribution.

For this question, all of the variables occur only once, with no variable occurring more than once. Hence, there is no mode for this dataset.

d) Midrange is the arithmetic mean of the highest and lowest number in the dataset.

Mathematically,

Midrange = (Highest + Lowest)/2

Midrange = (90 + 78)/2

Midrange = 84

e) Range is the difference the highest and the lowest numbers in a dataset.

Range = 90 - 78 = 12

f) Variance is an average of the squared deviations from the mean.

Mathematically,

Variance = [Σ(x - xbar)²/N]

xbar = mean

Σ(x - xbar)² = (78 - 84.14)² + (80 - 84.14)² + (83 - 84.14)² + (85 - 84.14)² + (86 - 84.14)² + (87 - 84.14)² + (90 - 84.14)² = 102.8572

Variance = (102.8572)/7

Variance = 14.69

g) Standard deviation = √(variance)

Standard deviation = √(14.69)

Standard deviation = 3.83

Hope this Helps!!!

8 0

3 years ago

Read 2 more answers

Much power does a 10 Ohm bulb have in a series circuit with a battery of 12V and nothing else in the the circuit?￼￼

Much power does a 10 Ohm bulb have in a series circuit with a battery of 12V and nothing else in the the circuit?