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3 years ago

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Under most conditions, except ________ change, the resistance of an object is a constant and does not depend on the amount of th

e voltage or the amount of current passing through it.

1 answer:

Readme [11.4K]3 years ago

4 0

Answer:

Temperature change

Explanation:

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A 4-79 permalloy solenoid coil needs to produce a minimum inductance of 1.1 . If the maximum allowed current is 4 , how many tur

daser333 [38]

The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,

The magnetic field H is given as,

$H = \frac{nI}{l}$

Here,

n = Number of turns of the coil

I = Current that flows in the coil

l = Length of the coil

From the above equation, the number of turns of the coil is,

$n = \frac{Hl}{I}$

The magnetic field is again given by,

$H = \frac{B}{\mu_t}$

Where the minimum inductance produced by the solenoid coil is B.

We have to obtain n, that

$n = \dfrac{\frac{B}{\mu_t}l}{I}$

Replacing with our values we have that,

$n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}$

$n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}$

$n = 27.5 \approx 28$

Therefore the number of turn required is 28Truns

4 0

3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago

How many moles of He atoms are in 6.00 g of He?

S_A_V [24]

Answer:

Explanation:

9.5

4 0

3 years ago

Describe how the composition of gases changes as you travel up through Earth's atmosphere.

Ghella [55]

<span>As the gases go higher in the atmosphere they start to compose or join together</span>

7 0

3 years ago

Water flows through a cylindrical pipe of varying cross-section. The velocity is 3.0 m/s at a point where the pipe diameter is 1

Setler79 [48]

Answer:

The flow rate is $R =2.357 *10^{-4} \ m^3/s$

Explanation:

From the question we are told that

The velocity is $v = 3.0 \ m/s$

The diameter of the pipe is $d = 1.0 \ cm = 0.01 \ m$

The radius of the pipe is mathematically represented as

$r = \frac{d}{2}$

substituting values

$r = \frac{0.01}{2}$

$r = 0.005 \ m$

The flow rate is mathematically represented as

$R = v * A$

Where is the cross-sectional area of the pipe which is mathematically evaluated as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.005)^2$

$A = 7.855 * 10^{-5} \ m^2$

So

$R = 3.0 * 7.855 *10^{-5}$

$R = 2.357*10^{-4} \ m^3 /s$

8 0

3 years ago