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sasho [114]
3 years ago
11

A single loop of current is immersed in an externally applied uniform magnetic field of 3 Tesla oriented in the positive y direc

tion: begin mathsize 12px style B with rightwards arrow on top equals 3 space j with logical and on top end style T. The loop carries a 1/2 amp current, I = 0.5 A. If the lengths of the sides of the loop are 4 m and 2 m, what is the magnitude of the magnetic moment, mu, of the loop?
Physics
1 answer:
vlabodo [156]3 years ago
6 0

Answer:

mu=12Tm^2

Explanation:

the magnetic moment mu of a single loop is given by:

\mu = I A B

where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:

\mu=(0.5A)(4m*2m)(3T)=12Tm^2

hope this helps!!

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solniwko [45]
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr 
</span><span>Θ = arctan(v0² / gr) </span>

<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>

<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
4 0
2 years ago
A special rocket can produce 7.66 ✕ 10^5 N of instantaneous thrust with an exhaust speed of 3.05 ✕ 103 m/s in vacuum. What mass
adelina 88 [10]

The mass of fuel the engine burn each second to produce a thrust of 7.66×10⁵ N is 2.5×10² kg/s.

<h3 /><h3>What is mass?</h3>

Mass can be defined as the quantity of matter contained in a body. The S.I unit of mass is kilogram(kg)

To calculate the mass the engine burns each seconds, we use the formula below.

Formual:

  • M = T/v............. Equation

Where:

  • M = Mass per seconds of the rocket
  • T = Thrust
  • v = Velocity

From the question,

Given:

  • T = 7.66×10⁵ N
  • v = 3.05×10³ m/s

Substitute these values into equation 1

  • M = (7.66×10⁵)/(3.05×10³)
  • M = 2.5×10² kg/s

Hence, the mass of fuel burned in each second is 2.5×10² kg/s.

Learn more about mass here: brainly.com/question/25121535

#SPJ1

8 0
1 year ago
Mechanical energy is the sum of ________ energy and potential energy.apex
crimeas [40]
Mechanical energy is the sum of kinetic energy and potential energy
6 0
3 years ago
Help me with this physics math?
solong [7]
Ok. PEMDAS tells us to take care of the square first. When we do that, the denominator becomes

(6.4)^2 x 10^12

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Now it's just a matter of mashing out the fraction.

The 'mantissa' (the number part) is

6/40.96 = 0.1465

and the order of magnitude is

10^24 / 10^12 = 10^12 .

Put it all together and you've got

1.465 x 10^11 .
4 0
3 years ago
What is the speed of a wave that has a frequency of 125 Hz and a wavelength of 1.25 meters? Express your answer to the nearest w
shtirl [24]

156 is the answer. so 156.25 is almost the same thing, you just round. It's not hard. Thank you!!!

3 0
3 years ago
Read 2 more answers
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