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nadezda [96]
3 years ago
10

A 0.49 m copper rod with a mass of 0.15 kg carries a current of 13 A in the positive y direction. Let upward be the positive dir

ection. What is the magnitude of the minimum magnetic field needed to levitate the rod
Physics
1 answer:
Helen [10]3 years ago
6 0

Answer:

Magnetic field, B = 0.23 T          

Explanation:

Given that,

Length of the copper rod, L = 0.49 m

Mass of the copper rod, m = 0.15 kg

Current in rod, I = 13 A (in +ve y direction)

When the rod is placed in magnetic field, the magnetic force is balanced by its weight such that :

BIL=mg\\\\B=\dfrac{mg}{IL}\\\\B=\dfrac{0.15\times 9.8}{13\times 0.49}\\\\B=0.23\ T

So, the magnitude of the minimum magnetic field needed to levitate the rod is 0.23 T.

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An astronaut landed on a far away planet that has a sea of water. To determine the gravitational acceleration on the planet's su
TiliK225 [7]

The concept required to solve this problem is hydrostatic pressure. From the theory and assuming that the density of water on that planet is equal to that of the earth (1000kg / m ^ 3)we can mathematically define the pressure as

P = \rho g h

Where,

\rho = Density

h = Height

g = Gravitational acceleration

Rearranging the equation based on gravity

g = \frac{P_h}{\rho h}

The mathematical problem gives us values such as:

P = 2.4 atm (\frac{101325Pa}{1atm}) = 243180Pa

\rho = 1000kg/m^3

h = 28.6m

Replacing we have,

g = \frac{243180}{(1000)(28.6)}

g = 8.5m/s^2

Therefore the gravitational acceleration on the planet's surface is 8.5m/s^2

3 0
3 years ago
Consider a household that uses 22.0 kW-hour of electricity per day on average. Most of that electricity is supplied by fossil fu
Lelu [443]

Answer:

A_{net} = 13.469\,m^{2}

Explanation:

The daily energy consumption of the household is:

\Delta E_{day} = (22\,kWh)\cdot (\frac{3600000\,J}{1\,kWh} )

\Delta E_{day} = 79200000\,J

The needed area of solar panel to power the household is:

\Delta E_{day} = \eta\cdot G\cdot A_{net}\cdot \Delta t

A_{net} = \frac{\Delta E_{day}}{\eta\cdot G\cdot \Delta t}

A_{net} = \frac{79200000\,J}{(0.181)\cdot (376\,\frac{W}{m^{2}} )\cdot (86400\,s)}

A_{net} = 13.469\,m^{2}

3 0
3 years ago
Using the balanced chemical equation below, calculate how many moles of ammonia would be produced when 9.1 moles of hydrogen gas
strojnjashka [21]

6.07 moles

Explanation:

Given parameters:

Number of moles of reacting hydrogen gas = 9.1 moles

Unknown:

Number of moles of ammonia produced = ?

Solution:

  Balanced chemical equation:

           N₂   +     3H₂    →      2NH₃

We should work from the known to the unknown specie. We known the number of moles of hydrogen gas reacting. We simply relate this to that of the ammonia.

 From the reaction:

   3 mole of hydrogen gas produced 2 mole of ammonia

     9.1 moles of the hydrogen gas will produce:  \frac{2 x 9.1}{3}

                                                                            = 6.07 moles

Learn more:

Number of moles brainly.com/question/1841136

#learnwithbrainly

3 0
3 years ago
How is 0.00069 written in scienitfic notation?
victus00 [196]
You need to move the decimal point between the six and nine. 6.9 X 10^-4
8 0
3 years ago
Convert 400 mm to m using the method of dimensional analysis
s2008m [1.1K]

Answer:

To convert 400 mm to m you can apply the formula [m] = [mm] / 1000; use 400 for mm. Thus, the conversion 400 mm m is the result of dividing 400 by 1000. 0.4

<em>PLEASE</em><em> </em><em>MARK</em><em> </em><em>AS</em><em> </em><em>BRAINLIEST</em><em> </em><em>ANSWER</em><em> </em>

7 0
2 years ago
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