Answer:
Magnetic dipole moment is 0.0683 J/T.
Explanation:
It is given that,
Length of the rod, l = 7.3 cm = 0.073 m
Diameter of the cylinder, d = 1.5 cm = 0.015 m
Magnetization, 
The dipole moment per unit volume is called the magnetization of a magnet. Mathematically, it is given by :
Where
r is the radius of rod, r = 0.0075 m
So, its magnetic dipole moment is 0.0683 J/T. Hence, this is the required solution.
Answer:
prinsip bernoulli
Explanation:
Prinsip Bernoulli adalah sebuah istilah di dalam mekanika fluida yang menyatakan bahwa pada suatu aliran fluida, peningkatan pada kecepatan fluida akan menimbulkan penurunan tekanan pada aliran tersebut. Prinsip ini sebenarnya merupakan penyederhanaan dari Persamaan Bernoulli yang menyatakan bahwa jumlah energi pada suatu titik di dalam suatu aliran tertutup sama besarnya dengan jumlah energi di titik lain pada jalur aliran yang sama. Prinsip ini diambil dari nama ilmuwan Belanda/Swiss yang bernama Daniel Bernoulli.
The average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.
<h3>How to calculate the Average speed?</h3>
The average speed can be calculated by adding the speed of each trial divided by the number of trials,
For 0.25 m the average speed will be:
For the 0.50 m, the average speed will:
Therefore, the average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.
Learn more about Average speed:
brainly.com/question/26386984
Answer:
I just answered a question like this. It should be B. An electromagnet :)
Explanation:
Answer:
The thermal conductivity of the wall = 40W/m.C
h = 10 W/m^2.C
Explanation:
The heat conduction equation is given by:
d^2T/ dx^2 + egen/ K = 0
The thermal conductivity of the wall can be calculated using:
K = egen/ 2a = 800/2×10
K = 800/20 = 40W/m.C
Applying energy balance at the wall surface
"qL = "qconv
-K = (dT/dx)L = h (TL - Tinfinity)
The convention heat transfer coefficient will be:
h = -k × (-2aL)/ (TL - Tinfinty)
h = ( 2× 40 × 10 × 0.05) / (30-26)
h = 40/4 = 10W/m^2.C
From the given temperature distribution
t(x) = 10 (L^2-X^2) + 30 = 30°
T(L) = ( L^2- L^2) + 30 = 30°
dT/ dx = -2aL
d^2T/ dx^2 = - 2a
