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Jlenok [28]
3 years ago
8

Settlers are trying to determine the value of g on a distant planet. They throw a wrench downward from a height of 3 m. If the i

nitial speed of the wrench is 2 m/s, and the speed right before impact is 10 m/s, what is the value of g?
Physics
1 answer:
Agata [3.3K]3 years ago
8 0
<h2>Value of g on the distant planet is 16 m/s²</h2>

Explanation:

We have equation of motion v² = u² + 2as

Here initial velocity, u = 2 m/s

Final velocity, v = 10 m/s

Displacement, s = 3 m

We need to find acceleration, a.

Substituting

               v² = u² + 2as

               10² = 2² + 2 x a x 3

                6a =  96

                  a = 16 m/s²

Value of g on the distant planet is 16 m/s²

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Suppose that instead of a long straight wire, a shortstraight wire was used. The distance from the wire to thepoint that the mag
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Answer:

Thus, if field were sampled at same distance, the field due to short wire is greater than field due to long wire.

Explanation:

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As the resistance of the long cable is proportional to the cable length, the short cable becomes less resilient than the long cable, so going through the short cable (where filled with the same material) is a bigger amount of currents. If the field is measured at the same time, the field is therefore larger than the long wire because of the short wire.

4 0
3 years ago
Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23
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}}

=>    \frac{T_1}{T_2}  =  [\frac{23500 }{9380} ]^{\frac{3}{2}}]

=>    \frac{T_1}{T_2}  = 3.965

   

8 0
3 years ago
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