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Nutka1998 [239]
2 years ago
3

An insulated wire of diameter 1.0 mm and negligible resistance is wrapped tightly around a cylindrical core of radius 5.0 cm and

length 30 cm to build a solenoid. What is the energy stored in this solenoid when a current I = 0.20 A flows through it? (μ0 = 4π × 10-7 T ∙ m/A)
Physics
1 answer:
zmey [24]2 years ago
7 0

Answer:

5.92 x 10⁻⁵ J

Explanation:

N  = Total number of turns

d  = diameter of insulated wire = 1 mm = 0.001 m

l  = length of the cylindrical core = 30 cm = 0.30 m

Length is given as

l = N d

0.30 = N (0.001)

N  = 300

r = radius of the core = 5 cm = 0.05 m

Area is given as

A = πr²

A = (3.14) (0.05)²

A = 0.00785 m²

Inductance of the solenoid is given as

L = \frac{\mu _{o}N^{2}A}{l}

L = \frac{(12.56\times 10^{-7}))(300)^{2}(0.00785)}{0.30}

L = 0.00296

i = current in the solenoid = 0.20 A

Energy stored in the solenoid is given as

U = (0.5)Li^{2}

U = (0.5)(0.00296)(0.20)^{2}

U = 5.92 x 10⁻⁵ J

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Answer:

  0.00325 moles/liter/second

Explanation:

The tangent line has a slope of (y2 -y1)/(x2 -x1) = (0.35-0.48)/(40-0) = -0.00325.

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2 years ago
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Sloan [31]

Answer:

The ratio is   \frac{T_1}{T_2}  = 3.965

Explanation:

From the question we are told that

   The  radius of Phobos orbit is  R_2 =  9380 km

    The radius  of Deimos orbit is  R_1  =  23500 \  km

Generally from Kepler's third law

    T^2 =  \frac{ 4 *  \pi^2 *  R^3}{G * M  }

Here M is the mass of Mars which is constant

        G is the gravitational  constant

So we see that \frac{ 4 *  \pi^2  }{G * M  } =  constant

   

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=>  [\frac{T_1}{T_2} ]^2 =  [\frac{R_1}{R_2} ]^3

Here T_1 is the period of Deimos

and  T_1 is the period of  Phobos

So

      [\frac{T_1}{T_2} ] =  [\frac{R_1}{R_2} ]^{\frac{3}{2}}

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8 0
2 years ago
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v_i=-43 m/s (with negative sign since it is due south)
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4 0
2 years ago
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Helga [31]

Answer:

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v is the velocity after impact

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Required

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v = 41.27m/s

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8 0
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Rashid [163]

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