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3 years ago

10

the primary functionn of the bias circuit is to hold the circuit stable at the destinated Q-point. true or false

1 answer:

DIA [1.3K]3 years ago

3 0

Shake r risks e wish sis false

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We have a long wire with a circular cross section and radius a = 2.40 cm. The current density in this wire is uniform, with a to

Alika [10]

Answer:

The magnitude of the magnetic field is 7.49x10⁻⁶T

Explanation:

The magnetic field is:

$B=(\frac{\mu i}{2\pi R^{2} } )r$

Where

i = current = 3 A

R = radius = 2.4 cm = 0.024 m

r = distance = 0.72 cm = 7.2x10⁻³m

Replacing:

$B=(\frac{4\pi x10^{-7} *3}{2\pi *0.024^{2} } )*7.2x10^{-3}= 7.49x10^{-6} T$

7 0

3 years ago

Three identical very dense masses of 7500 kg each are placed on the x axis. One mass is at x1 = -100 cm , one is at the origin,

sukhopar [10]

Answer:

0.00354 (N)

Explanation:

Convert to metric system:

$x_1 = -100 cm = 1 m$

$x_2 = 420 cm = 4.2 m$

Formula for gravitational force:

$F_g = G\frac{mM}{s^2}$

where s is the distance between 2 bodies masses m and M

Substitute the number to the formula above and since the 2 forces are acting in opposite direction, the total net gravitational force on the mass of origin be:

$F_g = F_{g1} - F_{g2}$

$F_g = G\frac{m_1M}{x_1^2} - G\frac{m_2M}{x_2^2}$

$F_g = GM(\frac{m_1}{x_1^2} - \frac{m_2}{x_2^2})$

$F_g = 6.67*10^{-11} * 7500 (\frac{7500}{1^2} - \frac{7500}{4.2^2})$

$F_g = 5*10^{-7}(7500 - 425.17)$

$F_g = 5*10^{-7} * 7074.83$

$F_g = 0.00354 (N)$

5 0

3 years ago

Please help on this one?

nikdorinn [45]

a is the right answer

4 0

3 years ago

Read 2 more answers

What is the speed of sound dependent on?

mojhsa [17]

Depends on the elasticity and density of the medium through what it is traveling <span> </span>

5 0

3 years ago

Block A can slide relative to block B which, in turn, can slide on a perfectly smooth horizontal plane. If the initial velocity

Anna11 [10]

Answer:

the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$

the distance that A slides relative to B is $S = \frac{v_o^2M}{2 \mu g (M+m)}$

Explanation:

From the diagram below;

acceleration of A relative to B is : $a = - ( \mu g + \frac{ \mu mg}{M})$

where

v = u + at

$0 = v_o + ( - \mu g - \frac{\mu m g }{M})t$

Making t the subject of the formula; we have:

$t = \frac{v_o M}{(\mu g )(M+m)}$

$v^2 = u^2 +2 as\\\\0^2 = v_o^2 - 2 (\mu g ) (\frac{M+m}{M})S\\\\$

$S = \frac{v_o^2M}{2 \mu g (M+m)}$ which implies the distance that A slides relative to B.

The final velocities of the two blocks can be determined as follows:

v = u + at

$v = v_o - \mu g \frac{v_oM}{\mu g (M+m)}\\\\v = \frac{\mu g mv_o}{m+M}\\\\$

$v = \frac{mv_o}{m+M}$

Thus, the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$

4 0

3 years ago