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babymother [125]
3 years ago
14

The acceleration due to gravity on Mercury is 3.7 m/s2, which is about six times the acceleration due to gravity on Pluto. Which

statement accurately compares the mass and weight of an object on these two planets?
An object weighs about one-sixth as much and has about one-sixth the mass on Pluto as on Mercury.
An object weighs about one-sixth as much but has the same mass on Pluto as on Mercury.
An object weighs the same but has about one-sixth the mass on Pluto as on Mercury.
An object weighs the same and has the same mass on Pluto as on Mercury.
Physics
2 answers:
ankoles [38]3 years ago
7 0

An object weighs about one-sixth as much but has the same mass on Pluto as on Mercury.

Explanation:

- The mass of an object is an intrinsed property of the object, and it tells the "amount of matter" inside the object. It does not depend on the location of the object.

- The weight of an object is equal to the force of gravity exerted by the planet on the object, and its magnitude is given by

W=mg (1)

where m is the object's mass and g is the gravitational acceleration on the planet.

From these two definitions, we clearly see that:

- the mass of an object is the same on Mars and on Pluto

- the weight of an object on Pluto is 1/6 of its weight on Mars, because the value of g on Pluto is 1/6 of the value of g on Mars, so in the formula (1) we get a factor 1/6 that is the reason behind the difference in weight.

Volgvan3 years ago
5 0
The second choice is correct.
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Answer:

a)  Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F ,  V = 3.4375 V ,

c)  U = 54.7 nJ ,  d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

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        Q = C V

         

can also be calculated using geometry consideration

        C = e or A / d

         

we reduce to the SI system

       A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

       d = 1.53 cm = 1.53 10⁻² m

we substitute

         Q = eo A / d V

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         Q = 3.9757 10⁻¹⁰ C

         

let's reduce to pC

         Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

          Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

           Q = k Q₀

           Q = 80 397.57

           Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

           C = k C₀

           C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

           C = 1.157 10⁻¹⁰ F

           V = Vo / k

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c) The stored energy is

             U = ½ C V²

              U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²

             U = 5.47 10⁻⁸ J

let's reduce to nJ

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d) energy after submerging

             U = ½ (kCo) (Vo / k) 2

             U = ½ Co Vo2 / k

             U = U₀ / k

             U = 54.7 / 80 nJ

              U = 0.68375 nJ

the energy change is

         ΔU = U₀ -U

          ΔU = 54.7 - 0.687375

           

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