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Lisa [10]
2 years ago
6

A very long half-cylinder of radius r carries a total current is uniformly distributed across the circumference. Determine the m

agnetic field along the central axis of the cylinder.
Physics
1 answer:
Rufina [12.5K]2 years ago
8 0

The magnetic field along the central axis of the cylinder will be <em>determined </em>using the formula given that all parameters are known

dBz=\frac{uoR^2di}{2r^3}

<h3>Magnetic field </h3>

Generally, A magnetic field is also referred to as a vector field that shows the magnetic presence on electric charges and magnetic materials.

The formula for magnetic field  is

Where

u=4\pi*10^{-7H/M}

Therefore, the magnetic field along the central axis of the cylinder will be determined using the formula given that all parameters are known

For more information on Magnets

brainly.com/question/7802337

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How is acceleration related to the force with mass is constant according to newtons second law of motion motion
Blababa [14]

Explanation:

, Newton's second law of motion can be formally stated as follows. The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net Force in the same direction as the net force and inversely proportional to the mass of the object

3 0
3 years ago
Two point charges, A and B, are separated by a distance of 19.0 cm . The magnitude of the charge on A is twice that of the charg
ankoles [38]

Answer:

QA = 19μC

QB = 9.5 μC

Explanation:

  • The force that each charge exerts on the other must obey Coulomb's Law, as follows:

       F_{AB} = \frac{k*Q_{A} * Q_{B}}{r_{AB}^{2}}  (1)

  • We know that the value of the magnitude of FAB  is 45.0 N, the distance between QA and  QB is 0.19 m, and that QA = 2*QB.
  • Replacing in (1), we can solve for QB, as follows:

      Q_{B} = \sqrt{\frac{F_{AB}*r_{AB} ^{2}}{2*k} } = \sqrt{\frac{45.0N*(0.19m) ^{2}}{2*9e9N*m2/C2} } = 9.5e-6 C  (2)

  • Since QA = 2*QB
  • ⇒ QA = 2* 9.5μC = 19.0 μC
  • ⇒ QB = 9.5μC
5 0
2 years ago
What is the major difference between the natural frequency and the damped frequency of oscillation.​
natita [175]

Answer:

This causes the amplitude of the oscillation to decay over time. The damped oscillation frequency does not equal the natural frequency. Damping causes the frequency of the damped oscillation to be slightly less than the natural frequency

6 0
2 years ago
PLEASE HELP!!
BartSMP [9]

Answer:A) 19.8 seconds

B) 467.5 metres

Explanation: 100km/hr = 27.78m/s

So, it covers 27.78 metres in one second

Take the time it would cover 550m as x

Therefore 27.78m = 1second

550m = x

Cross multiplying

27.78x = 550m

Divide both sides by the coefficient of x

x = 550÷27.78

x = 19.8seconds

B) distance = speed × time

Speed = 85km/hr = 23.61m/s

Time = 19.8seconds

Therefore,

Distance = 23.61 × 19.8

Distance = 467.5m

4 0
3 years ago
A snowball is rolling down a hill at 4.5 m/s and accumulating snow as it goes. Its diameter begins at 0.50 m and ends at the bot
Reil [10]
To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.

The formula for centripetal acceleration is:
Centripetal Acceleration = v squared divided by r

where:
v = velocity, m/s
r= radium, m

assuming the velocity does not change:

at the top of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 0.25 m
                                      = 81 m/s^2

at the bottom of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 1.25 m
                                      = 16.2 m/s^2

to find the change in centripetal acceleration, take the difference of the two.
change in centripetal acceleration = centripetal acceleration at the top of the hill - centripetal acceleration at the bottom of the hill

= 81 m/s^2 - 16.2 m/s^2
= 64.8 m/s^2 or 65 m/s^2
6 0
3 years ago
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