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4 years ago

5

PLZZ HELP WILL MARK BRAINLY The function of the respiratory system is to ___________.

1 answer:

pickupchik [31]4 years ago

3 0

Breathe and now I’m just filling in more letters so it’ll go thru

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A long, straight wire with a circular cross section of radius R carries a current I. Assume that the current density is not cons

igor_vitrenko [27]

Answer: (a) α = $\frac{3I}{2.\pi.R^{3}}$

(b) For r≤R: B(r) = μ_0. $(\frac{I.r^{2}}{2.\pi.R^{3}})$

For r≥R: B(r) = μ_0. $(\frac{I}{2.\pi.r})$

Explanation:

(a) The current I enclosed in a straight wire with current density not constant is calculated by:

$I_{c} = \int {J} \, dA$

where:

dA is the cross section.

In this case, a circular cross section of radius R, so it translates as:

$I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr$

$I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr$

$I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}$

$\alpha = \frac{3I}{2.\pi.R^{3}}$

For these circunstances, α = $\frac{3I}{2.\pi.R^{3}}$

(b) <u>Ampere's</u> <u>Law</u> to calculate magnetic field B is given by:

$\int\ {B} \, dl =$ μ_0. $I_{c}$

(i) First, first find $I_{c}$ for r ≤ R:

$I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr$

$I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr$

$I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr$

$I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}$

$I_{c} = \frac{I.r^{3}}{R^{3}}$

Calculating B(r), using Ampere's Law:

$\int\ {B} \, dl =$ μ_0. $I_{c}$

$B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} )$ .μ_0

B(r) = $(\frac{Ir^{3}}{R^{3}2.\pi.r})$ .μ_0

B(r) = $(\frac{Ir^{2}}{2.\pi.R^{3}} )$ .μ_0

For r ≤ R, magnetic field is B(r) = $(\frac{Ir^{2}}{2.\pi.R^{3}} )$ .μ_0

(ii) For r ≥ R:

$I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr$

So, as calculated before:

$I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}$

$I_{c} =$ I

Using Ampere:

B.2.π.r = μ_0.I

B(r) = $(\frac{I}{2.\pi.r} )$ .μ_0

For r ≥ R, magnetic field is; B(r) = $(\frac{I}{2.\pi.r} )$ .μ_0.

3 0

3 years ago

The work done in holding a 30kg of bricks to a height of 20m is

ella [17]

The work done to "HOLD" a load of bricks at any height is zero.
Work is done only when force acts through a DISTANCE.

The work done to LIFT 30 kg of anything to 20m higher than
it already is, is

(force) · (vertical distance)

= (mass) · (gravity) · (vertical distance)

= (30 kg) · (9.8 m/s²) · (20 m)

= 5,880 joules

8 0

3 years ago

A timeline is necessary a. For every goal c. For every life goal b. For every academic goal d. For every high school goal please

kiruha [24]

I’d say for every life goal

5 0

3 years ago

Read 2 more answers

A spring with a spring constant of 165 N/m is attached to a 2.0 kg mass and set into motion.

marin [14]

Explanation:

We have,

Spring constant of the spring, k = 165 N/m

Mass, m = 2 kg

It is required to find the period of the mass-spring system. For the spring mass system, the period is given by :

$T=2\pi \sqrt{\dfrac{m}{k} }\\\\T=2\pi \sqrt{\dfrac{2}{165} }\\\\T=0.69\ s$

The frequency of vibration is reciprocal of its time period. So,

$f=\dfrac{1}{T}\\\\f=\dfrac{1}{0.69}\\\\f=1.44\ Hz$

So, the period of the mass-spring system is 0.69 s and frequency is 1.44 Hz.

3 0

4 years ago

When Jacob bats a baseball with a net force of 4.719 N, it accelerates 33 m/s2. What is the mass of the baseball?

Natasha2012 [34]

Answer:

its 24

Explanation:

5 0

3 years ago