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geniusboy [140]
3 years ago
10

If gravity on the moon is 1.62m/s, what is the weight of a 200g rock on the moon

Physics
1 answer:
igor_vitrenko [27]3 years ago
5 0
324 because when we use the formula for weight which is w=mg; W is weight; M is mass; G is gravity, now we multiply the gravity which is 1.62 m/s times the mass which is 200g and we will have 324. Hope it helps
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Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time–when they\'re not sleep
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Answer:

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Explanation:

To calculate the moment of the mice we must multiply their mass by their velocities, remember that the moment is a vector quantity, so we use the components of velocity

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  Px = m Vx

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Mouse 2

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 Px2 = 0.0223 (-0.883) = -0.0196 Kg m

 Py2 = 0.0223 (-0.253) = -0.00564 Kg m

 P2 = (-0.0196, -0.00564) Kg m

Mouse 3

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 P3 = (0.00680, 0.0158) Kg m

Mouse4

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  Px4 = 0.0127 (-0.555) = -0.00705 Kg m

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  P4 = (-0.00705, 0.00260) Kg m

To find the total momentum we must add each component of the individual moments

   Px = Px1 + Px2 + Px3 + Px4  

   Py = py1 + Py2 + Py3 + Py4

   Px = 0.0196 -0.0196 +0.00680 -0.00705

   Px = -0,000250 Kg m

   Py = -0.00637 -0.00564 +0.0158 +0.00260

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   P = (-0.000250, 0.00639) Kg m

7 0
3 years ago
A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu
Montano1993 [528]

Answer:

The angle of separation is  \Delta \theta =  0.93 ^o

Explanation:

From the question we are told that

    The angle of incidence is  \theta  _ i  = 56^o

     The refractive index of violet light  in diamond  is  n_v = 2.46

       The refractive index of red light in diamond is n_r = 2.41

      The wavelength of violet light is  \lambda _v = 400nm = 400*10^{-9}m

         The wavelength of red  light is  \lambda _r = 700nm = 700*10^{-9}m

Snell's  Law can be represented mathematically as

         \frac{sin \theta_i}{sin \theta_r} = n

Where \theta_r is the angle of refraction

=>       sin \theta_r  =   \frac{sin \theta_i}{n}

Now considering violet light

               sin \theta_r__{v}}  =   \frac{sin \theta_i}{n_v}

substituting values

                sin \theta_r__{v}}  =   \frac{sin (56)}{2.46}

                 sin \theta_r__{v}}  =  0.337

                 \theta_r__{v}}  =  sin ^{-1} (0.337)

                 \theta_r__{v}}  =  19.69^o

Now considering red light

               sin \theta_r__{R}}  =   \frac{sin \theta_i}{n_r}

substituting values

                sin \theta_r__{R}}  =   \frac{sin (56)}{2.41}

                 sin \theta_r__{R}}  =  0.344

                 \theta_r__{R}}  =  sin ^{-1} (0.344)

                 \theta_r__{R}}  = 20.12^o

The angle of separation between the red light and the violet light is mathematically evaluated as

                  \Delta \theta = \theta_r__{R}} -  \theta_r__{V}}

substituting values

                  \Delta \theta =20.12 - 19.69

                  \Delta \theta =  0.93 ^o

6 0
3 years ago
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