YLCCIC TPENTARS - Unscramble

How many minutes till 8:5 to 12:50

melamori03 [73]

Answer:

4:10

Explanation:

4 0

3 years ago

Can a constant magnetic field set into motion an electron initially at rest? Explain your answer.

Vinvika [58]

No, a constant magnetic field cannot set an electron initially at rest into motion

A force that accelerates a particle is necessary to change its velocity. The magnetic force is inversely proportional to the particle's speed. There cannot be a magnetic force acting on a moving particle, according to Einstein. A flux is a precise description of the greater-than-unity magnetic determine involving energy currents and magnet resources. The magnetic flux in a stage is actually selected apart from each some sort of route and also a degree (or durability); therefore, it is just a vector industry. The magnetic flux is usually defined as the Lorentz force that acts on moving galvanic costs.

To know more about Lorentz force refer to brainly.com/question/15552911

#SPJ4

5 0

2 years ago

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) Ampere's Law 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

If the average velocity of a duck is zero in a given time interval, what can you say about the displacement of the duck for that

Gemiola [76]

Answer:

Also 0

Explanation:

Since average speed of an object is that object's displacement over a unit of time, when an average speed is 0, its displacement over a unit of time must also be 0. When an average speed is not 0, then its displacement over a unit of time is also not 0 for that interval.

6 0

3 years ago

A 30000 grams boy is riding a merry-go-round with a radius of 600 cm. What is the centripetal force and acceleration on the boy

irga5000 [103]

Answer:

centripetal force is calculated by mass(kg) × tangetial velocity(m/s) ÷ radius (m)

Explanation:

so 30000g= 30kg

50km/h = 13.88m/s

600cm= 6m

30×13.88÷6= 69.4N

N= Newton's

hope this helps.

btw I'm 16 and love physics so I tried my best in this hope it went well!!

7 0

3 years ago