Ask question

Ask question

All categories

2 years ago

14

I am confused. Magnetic flux linkage and magnetic flux don't have the same definition, but they have same units. But there formu

lae are different, what are they?

1 answer:

ss7ja [257]2 years ago

6 0

Answer:

magnetic flux linkage includes number of turns, N while magnetic flux has no N

Since number of turns are not significant to units, they become similar.

You might be interested in

Motion equation of object is

Ganezh [65]

Take the derivative to find the velocity of the object:

$v=\dfrac{\mathrm dx}{\mathrm dt}=-12+6t$

The object stops when $v=0$ :

$-12+6t=0\implies6t=12\implies t=2$

so the answer is E.

5 0

3 years ago

at room temperature, iron is a solid and mercury is a liquid. based on this information we can infer thata. iron has a higher bo

olchik [2.2K]

A. iron has a higher boiling point thanmercury.

8 0

3 years ago

Read 2 more answers

A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformati

Slav-nsk [51]

The question is incomplete. The complete question is :

A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformation cycle. At a time, t = 0, a tensile stress of 20 MPa is applied instantaneously and maintained for 100 s. The stress is then removed at a rate of 0.2 MPa s−1 until the polymer is unloaded. If the creep compliance of the material is given by:

J(t) = Jo (1 - exp (-t/to))

Where,

Jo= 3m^2/ GPA

to= 200s

Determine

a) the strain after 100's (before stress is reversed)

b) the residual strain when stress falls to zero.

Answer:

a)-60GPA

b) 0

Explanation:

Given t= 0,

σ = 20Mpa

Change in σ= 0.2Mpas^-1

For creep compliance material,

J(t) = Jo (1 - exp (-t/to))

J(t) = 3 (1 - exp (-0/100))= 3m^2/Gpa

a) t= 100s

E(t)= ΔσJ (t - Jo)

= 0.2 × 3 ( 100 - 200 )

= 0.6 (-100)

= - 60 GPA

Residual strain, σ= 0

E(t)= Jσ (Jo) ∫t (t - Jo) dt

3 × 0 × 200 ∫t (t - Jo) dt

E(t) = 0

5 0

3 years ago

A very long string (linear density 0.7 kg/m ) is stretched with a tension of 70 N . One end of the string oscillates up and down

rewona [7]

To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

$V = \sqrt{\frac{T}{\rho}}$

Where,

T = Tension

$\rho =$ Linear density

Our data are given by

Tension , T = 70 N

Linear density , $\rho = 0.7 kg/m$

Amplitude , A = 7 cm = 0.07 m

Period , t = 0.35 s

Replacing our values,

$V = \sqrt{\frac{T}{\rho}}$

$V = \sqrt{\frac{70}{0.7}$

$V = 10m/s$

Speed can also be expressed as

$V = \lambda f$

Re-arrange to find \lambda

$\lambda = \frac{V}{f}$

Where,

f = Frequency,

Which is also described in function of the Period as,

$f = \frac{1}{T}$

$f = \frac{1}{0.35}$

$f = 2.86 Hz$

Therefore replacing to find $\lambda$

$\lambda = \frac{10}{2.86}$

$\lambda = 3.49m$

Therefore the wavelength of the waves created in the string is 3.49m

3 0

3 years ago

A 3.0-kg block moves up a 40o incline with constant speed under the action of a 26-N force acting up and parallel to the incline

CaHeK987 [17]

Answer:

it is very hard question for me sorry i cant solve it

3 0

3 years ago