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Rainbow [258]
3 years ago
15

The moon Ariel orbits Uranus at a distance of 1.91 x 108 m once every 2.52 days. Use that data to calculate the mass of Uranus.

[?] x 10kg​
Physics
1 answer:
PIT_PIT [208]3 years ago
4 0

Mu = 8.66 × 10^25 kg

Explanation:

centripetal force = gravitational force

m \frac{ {v}^{2} }{r}  = (grav.const) \frac{m \times mu}{ {r}^{2} }

where

m = mass of moon Ariel

mu = mass of Uranus

r = radius of Ariel's orbit

v = Ariel's velocity around Uranus

To find the velocity, we need to find the circumference of the no orbit and then divide it by the period (2.52 days):

circumference = 2πr = 2π×(1.91 × 10^8 m)

= 1.2 × 10^9 m

period = 2.52 days × (24 h/1 day)×(3600 s/1 hr)

= 2.18 × 10^5 s

v = (1.2 × 10^9 m)/(2.18 × 10^5 s)

= 5.5 × 10^3 m/s

(5.5 × 10^3 m/s)^2/(1.91 × 10^8 m) = (6.67 × 10^-11 m^3/kg-s^2)Mu/(1.91 × 10^8 m)^2

Solving Mu,

Mu = 8.66 × 10^25 kg

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3 years ago
Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2
Bess [88]

Answer:

The acceleration of M_2 is  a =  0.7156 m/s^2

Explanation:

From the question we are told that

    The mass of first block is  M_1 =  2.25 \ kg

    The angle of inclination of first block is  \theta _1 =  43.5^o

    The coefficient of kinetic friction of the first block is  \mu_1  = 0.205

      The mass of the second block is  M_2 = 5.45 \ kg

     The angle of inclination of the second block is  \theta _2 =  32.5^o

      The coefficient of kinetic friction of the second block is \mu _2 = 0.105

The acceleration of M_1 \ and\  M_2 are same

The force acting on the mass M_1 is mathematically represented as

     F_1 = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

=> M_1 a = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

Where T is the tension on the rope

The force acting on the mass M_2 is mathematically represented as    

  F_2 =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

   M_2 a =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

At equilibrium

  F_1 =  F_2

So

 T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

making a the subject of the formula

    a =  \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}

substituting values a =  \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}

    => a =  0.7156 m/s^2

     

3 0
3 years ago
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