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Juliette [100K]
4 years ago
15

4.A 31-cm long conducting wire of 9-g carrying 7- A current is placed in a uniform magnetic field. What are the strength and dir

ection of the magnetic field needed to levitate the wire?
Physics
1 answer:
Anika [276]4 years ago
7 0

Answer

Given,

Length of the wire,L = 31 cm = 0.31 m

mass of the wire, m = 9 g = 0.009 Kg

Current in the wire,I = 7 A

Magnetic field strength, B= ?

Equating magnetic force to the weight of the wire.

BIL = m g

B=\dfrac{m g}{IL}

B=\dfrac{0.009\times 9.81}{7\times 0.31}

B = 0.0407 T

For Force to be upward magnetic field direction should be outward of the plane of paper.

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Dryer sheets stop static cling by absorbing static electricity formed by different fabrics during your dryer cycle. The fabric softeners that coat dryer sheets are positively charged ions to balance the electrons and ions that cause static cling, leaving you with soft clothes without the static.

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The vacuum tube was replaced with what electrical component?
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The vacuum tube was replaced with transistor.

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3 years ago
Convert the number from scientific into standard notation: 5.9 x 10-2
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Move the decimal point to:
Left : (if the exponent of ten is a negative number -) ... OUR CASE HERE (-2)
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Do not write the power of ten anymore.

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Calculations made using Celsius or Fahrenheit will not work for gas law calculations
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4 years ago
The temperature of air changes from 0 to 18°C while its velocity changes from zero to a final velocity, and its elevation change
irina [24]

Answer:

For the air:

Final Velocity 160.77m/s

Final Elevation 1,317.43m

the Internal, Kinetic, and Potential Energy changes  will be equal.

Explanation:

In principle we know the following:

  • <u>Internal Energy:</u> is defined as the energy contained within a system (in terms of thermodynamics). It only accounts for any energy changes due to the internal system (thus any outside forces/changes are not accounted for). In S.I. is defined as U=mC_{V}\Delta T where m is the mass (kg), C_{V} is a specific constant-volume (kJ/kg°C) and \Delta T is the Temperature change in °C.
  • <u>Kinetic Energy:</u> denotes the work done on an object (of given mass m) so that the object at rest, can accelerate to reach a final velocity. In S.I. is defined as K=\frac{1}{2}mv^2 where v is the velocity of the object in (m/s).
  • <u>Potential Energy:</u> denotes the energy occupied by an object (of given mass m) due to its position with respect to another object. In S.I. is defined as P=mgh, where g is the gravity constant equal to 9,81m/s^2 and h is the elevation (meters).

<em>Note: The Internal energy is unaffected by the Kinetic and Potential Energies.</em>

<u>Given Information:</u>

  • Temperature Change 0°C → 18°C ( thus \Delta T=18°C )
  • Object velocity we shall call it v_{o} and v_{f}, for initial and final, respectively. Here we also know that v_{o}=0m/s^2
  • Object elevation we shall call it h_{o} and h_{f}, for initial and final, respectively. Here we also know that h_{o}= 0m

∴<em> We are trying to find v_{f} and h_{f} of the air where U, K and P are equal.</em>

Lets look at the change in Energy for each.

<u>Step 1: Change in Kinetic Energy=Change in Internal Energy</u>

\Delta E_{K}=\Delta U\\\frac{1}{2}m{v_{f}}^2- \frac{1}{2}m{v_{o}}^2=mC_{V}\Delta T

Here we recall that v_{o}=0m/s^2 and mass m is the same everywhere. Thus we have:

\frac{1}{2}m{v_{f}}^2=mC_{V}\Delta T    

\frac{1}{2} {v_{f}}^2=C_{V}\Delta T\\ {v_{f}}^2=2C_{V}\Delta T\\ v_{f}=\sqrt{2C_{V}\Delta T}     Eqn(1)

<u>Step 2: Change in Potential Energy=Change in Internal Energy</u>

\Delta E_{P}=\Delta U\\mgh_{f}-mgh_{o}=mC_{V}\Delta T

Here we recall that h_{o}=0m/s^2 and mass m is the same everywhere. Thus we have:

mg(h_{f}-h_{o})=mC_{V}\Delta T\\gh_{f}=C_{V}\Delta T\\

h_{f}=\frac{C_{V}\Delta T}{g}      Eqn(2).

Finally by plugging the known values in Eqns (1) and (2) we obtain:

v_{f}=\sqrt{2*718*18}=160.77m/s

h_{f}=\frac{718*18}{9.81}=1,317.43m

Thus we can conclude that for the air final velocity v_{f}=160.77m/s and final elevation h_{f}=1,317.43m the internal, kinetic, and potential energy changes  will be equal.

3 0
3 years ago
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