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3 years ago

12

Please helo me to 1st question

1 answer:

k0ka [10]3 years ago

6 0

Explanation:

6 min in seconds:

6×60

<h2><u>360</u><u> </u><u>s</u><u>e</u><u>c</u><u>o</u><u>n</u><u>d</u><u>s</u><u> </u></h2><h2 /><h2 />

6 min in hours:

<h2><u>0</u><u>.</u><u>1</u><u> </u><u>hours</u></h2>

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3 years ago

HELP ME PLEASE!!!!!!!!!

Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

$F = \frac{kq_1q_3}{d_{13}^2}$

now from the above equation

$F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}$

$F = 0.288 N$

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

$F_{net} = 2Fcos\theta$

here we know that angle is

$\theta = 37 degree$

now we have

$F_{net} = 2\times 0.288 cos37$

$F_{net} = 0.46 N$

so net force on q3 is 0.46 N vertically upwards along +Y axis

6 0

3 years ago

A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,

gavmur [86]

Answer:

a. ρ $_\beta=1.996J/m^3$

b. $U_E=9.445x10^{-15} J/m^3$

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

$i=14A$ , $d=2.5mm$ , $R=3.3$ Ω, $l=1 km$ , $E_o=8.85x10^{-12}F/m$ , $u_o=4*x10^{-7}H/m$

ρ $_\beta=\frac{\beta^2}{2*u_o}$

ρ $_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2$

ρ $_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}$

ρ $_\beta=1.996J/m^3$

b. The electric field can be find using the equation:

$U_E=\frac{1}{2}*E_o*E^2$

$E=(\frac{i*R}{l})^2$

$U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2$

$U_E=9.445x10^{-15} J/m^3$

4 0

3 years ago