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ikadub [295]
3 years ago
13

A planet orbits a star, in a year of length 3.37 x 107 s, in a nearly circular orbit of radius 1.04 x 1011 m. With respect to th

e star, determine (a) the angular speed of the planet, (b) the tangential speed of the planet, and (c) the magnitude of the planet's centripetal acceleration.
Physics
1 answer:
PIT_PIT [208]3 years ago
8 0

Answer

Given,

Time period of star,T = 3.37 x 10⁷ s

Radius of circular orbit,R = 1.04 x 10¹¹ m

a) Angular speed of the planet

   \omega = \dfrac{2\pi}{T}=\dfrac{2\pi}{3.37\times 10^{7}}

   \omega = 1.864\times 10^{-7}\ rad/s

b) tangential speed

   v = r \omega = 1.04\times 10^{11}\times 1.864 \times 10^{-7}

       v = 1.94 x 10⁴ m/s

c) centripetal acceleration magnitude

      a = \dfrac{v^2}{r}= \dfrac{(1.94\times 10^4)^2}{1.04\times 10^{11}}

          a = 3.62 x 10⁻³ m/s²

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3 years ago
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a train is moving with an initial velocity of 30 m/s, the brakes are applied so as to produce a uniform acceleration of -1.5 m/s
Pepsi [2]

Answer:

\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

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To Find:

Time in which train will come to rest (t).

Explanation:

\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a:  \\  \sf \implies 0 = 30 + ( - 1.5)(t) \\   \sf  \implies 0 = 30 - 1.5(t) \\  \sf \implies 30 - 1.5(t) = 0 \\  \\  \sf Subtract  \: 30  \: from  \: both  \: sides: \\  \sf \implies (30 -  \boxed{ \sf 30}) - 1.5(t) =  \boxed{ \sf  - 30} \\  \\  \sf 30 - 30 = 0 :  \\  \sf \implies  - 1.5(t) =  - 30 \\  \\  \sf Divide  \: both  \: sides \:  of \:  - 1.5(t) =  - 30 \: by \:  - 1.5 :  \\  \sf \implies  \frac{  - 1.5(t)}{ \boxed{ \sf - 1.5}}  =  \frac{ - 30}{ \boxed{ \sf -1.5 }}  \\  \\  \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}}  = 1 :  \\  \sf \implies t =  \frac{ - 30}{ - 1.5}  \\  \\   \sf  \frac{ - 30}{ - 1.5}  =  \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}}  = 20 :  \\  \sf  \implies t = 20 \: sec

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Frequency Table
City                       Frequency
Indianapolis              124
Saint Louis                416
Chicago                  1,225
Milwaukee                 435

Relative Frequency Table
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Indianapolis              124/2200 = 31/550
Saint Louis                416/2200 = 52/275
Chicago                  1,225/2200 = 49/88
Milwaukee                 435/2200 = 87/440
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