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sukhopar [10]
3 years ago
6

An object is 6.0cm in front of a converging lens with a focal length of 10cm .

Physics
1 answer:
Bess [88]3 years ago
8 0

Answer:  A) see attach file .

B) the iamgen is uprigth

C) the imagen is virtual

Explanation: see attach file.

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A proton moves with a speed of 1.00 x 106 m/s perpendicular to a magnetic field, B. As a result, the proton moves in circle of r
Softa [21]

Answer:

0.0109 m ≈ 10.9 mm

Explanation:

proton speed = 1 * 10^6 m/s

radius in which the proton moves = 20 m

<u>determine the radius of the circle in which an electron would move </u>

we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)

For proton :

Mp*V^2 / rp  = qp *VB   ∴  rp = Mp*V / qP*B    ---------- ( 1 )

For electron:

re = Me*V/ qE * B -------- ( 2 )

Next: take the ratio of equations 1 and 2

re / rp = Me / Mp                                 ( note: qE = qP = 1.6 * 10^-19 C )

∴ re ( radius of the electron orbit )

= ( Me / Mp ) rp

= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20

= ( 5.45 * 10^-4 ) * 20

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3 years ago
The potential energy of a particle experiencing a certain kind of force is given by U(x) = 2x + 8 x measured in joules (J), for
masha68 [24]

Answer:

Explanation:

Given

Potential Energy of a particle is given by U(x)=2x+\frac{8}{x}

For minimum or maximum Potential Energy differentiate U(x) w.r.t x

\frac{\mathrm{d} U}{\mathrm{d} x}=2-\frac{8}{x^2}

putting \frac{\mathrm{d} }{\mathrm{d} x}=0

2-\frac{8}{x^2}=0

x^2=\frac{8}{2}

x^2=4

x=\sqrt{4}=\pm 2

To know minimum value check \frac{\mathrm{d^2} U}{\mathrm{d} x^2}

at x=-2

\frac{\mathrm{d^2} U}{\mathrm{d} x^2}=\frac{16}{x^3}=-\frac{16}{8}=-2

so at x=-2 potential energy is minimum U=-8\ J

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