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ruslelena [56]
3 years ago
9

A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is

the planet's radius? Hint: Because planets are the shape of a sphere, the moment of inertia is I = (2/5)mr2.
A) 6.38 x 106 m
B) 2.55 × 106 m
C) 6.52 × 1012 m
D) 4.07 × 1013m
Physics
1 answer:
malfutka [58]3 years ago
4 0

Answer:  A) 6.38(10)^{6} m

Explanation:

The equation for the moment of inertia I of a sphere is:

I=\frac{2}{5}mr^{2} (1)

Where:

I=9.74(10)^{37}kg m^{2} is the moment of inertia of the planet (assumed with the shape of a sphere)

m=5.98(10)^{24}kg is the mass of the planet

r is the radius of the planet

Isolating r from (1):

r=\sqrt{\frac{5I}{2m}} (2)

Solving:

r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}} (3)

Finally:

r=6381149.077m \approx 6.38(10)^{6} m

Therefore, the correct option is A.

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A velocity of 20 \: km/h

Explanation:

Since the magnitude of the vector is equal to the magnitude of velocity, velocity of the 2 cm vector represents a velocity of $10\times 2= 20 \: km/h$.

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3 years ago
Un autocar que circula a 81 km/h frena uniformemente con una aceleración de -4,5 m/s2.
horsena [70]

Answer:

a) \Delta x=56.25 m

b) imagen adjunta

Explanation:

a) Primero debemos hacer la conversión de 81 km/h a m/s, esto es 22.5 m/s.

Ahora, usando la ecuacion cinemática, en un movimiento acelerado tenemos:

v_{f}^{2}=v_{0}^{2}+2a \Delta x

Queremos encontrar la posición hasta detenerse, osea vf = 0.

\Delta x=\frac{-v_{0}^{2}}{2a}

\Delta x=\frac{-22.5^{2}}{-2*4.5}

\Delta x=56.25 m

b) Para este caso el gráfico se encuentra adjunto.                                      

Espero que te sirva de ayuda!                                                                                                                                                                          

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8 0
3 years ago
A person pushes on a box with a force of 500N. If the box does not move, what is the force of static friction on the box?
Alexxx [7]

Answer:

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Explanation:

4 0
3 years ago
An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ
jenyasd209 [6]

Answer:

a) i = -9.63 cm ,    h ’= .0.24075 cm   erect

b)  i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

               1 / f = (n-1) (1 / R₁ - 1 / R₂)

                1 / f = (1.70 -1)) 1 / ∞ - 1/13)

                1 / f = 0.0538

                 f = - 18.57 cm

Now we can use the constructor equation

             1 / f = 1 / o + 1 / i

             1 / i = 1 / f - 1 / o

              1 / i = -1 / 18.57 -1/20

               1 / i = -0.1038 cm

               I = -9.63 cm

For the height of the

image let's use magnification

                 m = h '/ h = - i / o

                  h ’= -h i / o

                  h ’= - 0.5 (-9.63) / 20

                  h ’= .0.24075 cm

b) we invert the lens

The focal length is

             1 / f = (1.70 -1) (1/13 - 1 / int)

              1 / f = 0.0538

             f = 18.57 cm

             1 / i = 1 / f -1 / o

             1 / I = 1 / 18.57 - 1/20

             1 / I = 3.85 10-3

             i = 259.74 cm

     

            h ’= - 0.5 259.74 / 20

             h ’= 6.4935 cm

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