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3 years ago

HELLPPPPpppppppp mMmmeeeeeee

Physics

1 answer:

sweet-ann [11.9K]3 years ago

5 0

Student B is correct.

(3Ω + 6Ω + 17Ω ) = 26Ω

12v / 26Ω = 0.46 Amperes

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1.14 Which of the following is an example of a force without touching? A A boy pushing a trolley. B The mass of a car. C с A mag

andriy [413]

Explanation:

Option C a magnet attracting a nail

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3 years ago

50POINTS PLEASE HELP SPACE QUESTION

leonid [27]

Regardless of the time of year, the northern and southern hemispheres always experience opposite seasons. This is because during summer or winter, one part of the planet is more directly exposed to the rays of the Sun than the other, and this exposure alternates as the Earth revolves in its orbit.

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A jet, sitting on the runway, takes off and accelerates at 8.0 m/s for 16s. How far did the jet travel down the runway?

nydimaria [60]

Answer:

2.4 m/s". 1

Explanation:

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3 years ago

What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J

Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

$I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)$

and on the surface the current is

$I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A$

(b)

For Jb = J0(1 - r/R)

$I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )$

and on the surface

$I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0$

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

$I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A$

and in a simmilar way for Jb

$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0

3 years ago

A student made the table shown to list some contact and non-contact forces. Examples of Forces Contact Forces Non-Contact Forces

aleksandr82 [10.1K]

Answer:

the second one is incorrect

Explanation:

I am smart and took the test

5 0

3 years ago