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3 years ago

PLEASE ANSWER

Physics

1 answer:

gtnhenbr [62]3 years ago

7 0

Answer:

<em><u>A balance and a beaker of water!</u></em>

Explanation:

Krishne wants to measure the mass and volume of a key. The tools that are to be used are – a balance and a beaker of water.

Mass will be represent with the amount of matter that is in an object. This will be measured in terms of kilogram. The mass can easily be measured with balance.

Volume is the quantity of matter that is in the object. This is measured in terms of cubic meter. If you drop the key in a beaker of water, the water inside the beaker will increase and this increases the volume of water that will be equal to volume of key.

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Explanation:

Given that,

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$F=6\pi \eta rv\\\\v=\dfrac{F}{6\pi \eta r}$

Put all the values,

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A transformer connected to a 120 V (rms) ac line is to supply 13,000 V (rms) for a neon sign. To reduce shock hazard, a fuse is

TiliK225 [7]

Answer with Explanation:

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5 0

4 years ago

Read 2 more answers

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