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Nostrana [21]
3 years ago
14

Which planet has the greatest mass?

Physics
2 answers:
Vesna [10]3 years ago
6 0

Answer: Jupiter

Explanation: The largest planet in our solar system is Jupiter and it has a mass that is 300 times Earth's mass.

Firlakuza [10]3 years ago
5 0

Answer:

Jupiter

Explanation:

The largest planet in our solar system by far is Jupiter, which beats out all the other planets in both mass and volume. Jupiter's mass is more than 300 times that of Earth, and its diameter, at 140,000 km, is about 11 times Earth's diameter.

Note:

Please mark as Brainliest! <3

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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

     

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3 years ago
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1. e) None of the above is necessarily true.

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An object starts at 4m/s and accelerates to 6m/s in 10 seconds. Find its displacement.
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Answer:

Explanation:

The 2 equations we need here are, first:

a=\frac{v_f-v_0}{t} and then once we solve for the acceleration here:

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Solving for acceleration:

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The answer to your problem is I believe (B.)
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