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Nikitich [7]
3 years ago
11

What is one major difference between the attributes of Mercury and Jupiter? A. Mercury is the planet that is closest to the Sun,

while Jupiter is the farthest. B. Mercury is the coldest planet in the solar system, while Jupiter is the hottest. C. Mercury has no orbiting satellites, while Jupiter has many satellites circling it. D. Mercury has bands and rings visible from Earth, while Jupiter has no such bands. E. Mercury has a rich supply of methane, while Jupiter has a rich supply of hydrogen.
Physics
2 answers:
Sergeu [11.5K]3 years ago
4 0

Answer:

C. Mercury has no orbiting satellites, while Jupiter has many satellites circling it.

Explanation:

Mercury is the smallest and closest planet to sun. It is a rocky planet. It does not have any natural satellites. On the other hand, Jupiter is the largest planet and is made up of gases. It has bands. Jupiter has more than 60 natural satellites revolving about it. It is mostly composed of hydrogen and helium.

Thus, the major difference between Mercury and Jupiter is C. Mercury has no orbiting satellites, while Jupiter has many satellites circling it.

sergeinik [125]3 years ago
3 0

The major difference is: Mercury has no orbiting satellites, while Jupiter has many satellites that surround it, so the correct answer is C

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An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 So

       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
3 years ago
A student performing a double-slit experiment is using a green laser with a wavelength of 550 nm. She is confused when the m = 5
sweet [91]

Answer:

d = 52 μm

Explanation:

given,

wavelength of the light source (λ)= 550 nm

distance to form interference pattern(D) = 1.5 m

y = 1.6 cm = 0.016 m

width of the slits = ?

now, using displacement formula

 y = \dfrac{m\lambda\ D}{d}

for the first maxima, m = 1

 d = \dfrac{1\times \lambda\ D}{y}

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hence, the width of her slits is equal to d = 52 μm

3 0
3 years ago
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Answer:

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4 0
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The final mass after decay can be obtained by using under given relation:

half life period of As-81 = 33 seconds

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= 100 x ( 1/2^(43.2/33))

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HACTEHA [7]
UV Radiation since it has a higher frequency than the others. The higher the frequency the shorter the wavelength.
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