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4 years ago

If the current density in a wire or radius R is given by J-k+5,0F wire? R, what is the current in the wire?

Physics

1 answer:

Deffense [45]4 years ago

8 0

Answer:

I = $R^{2}$ (K+5)

Explanation:

Given :

J = k+5

Now selecting a thin ring in the wire of radius "r" and thickness dr.

Current through the thin ring is

dI = J X 2πrdr

dI = (K+5) x 2πrdr

Now integrating we get

I = $\int_{0}^{R} = (K+5).2\pi rdr$

I = (K+5) 2π $\int_{0}^{R} rdr$

I = (K+5) 2π $\frac{R^{2}}{2}$

I = $R^{2}$ (K+5)

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Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez

WITCHER [35]

Answer:

Explanation:

a )

If it is totally absorbed pressure is calculated as follows .

Pressure = I / c where I is intensity of light falling .

= 1000 / 3 x 10⁸

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b ) weight of tritium atom

= 3 x 1.67 x 10⁻²⁷ kg

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= 3.33x 10⁻⁶ / 3 x 1.67 x 10⁻²⁷

= .6646 x 10²¹ m /s²

= 66.46 x 10¹⁹ m / s²

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3 years ago

A satellite is in orbit 36000km above the surface of the earth. Its angular velocity is 7.27*10^-5 rad/s. What is the velocity o

bagirrra123 [75]

Answer:

v = 3.08 km/s

Explanation:

Given that,

The angular velocity of the satellite = $\omega=7.27\times 10^{-5} rad/s$

A satellite is in orbit 36000km above the surface of the earth.

The radius of the earth is 6400 km

Let v is the velocity of the satellite. It can be calculated as :

$v=r\omega\\\\v=(36000\times 10^3+6400\times 10^3)\times 7.27\times 10^{-5}\\\\v=3082.48\ m/s\\\\v=3.08\ km/s$

So, the velocity of the satellite is 3.08 km/s.

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3 years ago

The additional product of the nuclear fission reaction shown in the

Arada [10]

The answer to your question is C.

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4 years ago

A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853

Norma-Jean [14]

Recall that

${v_f}^2-{v_i}^2=2a\Delta x$

where $v_i$ and $v_f$ are the initial and final velocities, respecitvely; $a$ is the acceleration; and $\Delta x$ is the change in position.

So we have

$\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)$

$\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}$

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

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3 years ago

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

3 0

4 years ago