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3 years ago

Need only group b 1st question

Physics

1 answer:

galina1969 [7]3 years ago

7 0

Answer:

49 m/s

Explanation:

This is the question from the document:

1. A bag dropped from a helicopter falls with an acceleration of 9.8m/s2. What is its velocity after 5s?

We can find the answer by using the equation a = (v-u) / t

Acceleration a = 9.81m/s2

initial velocity u = 0 m/s (the bag is only dropped, so there's no initial velocity already existing)

time t= 5s

final velocity v = unknown

9.8 = (v -0) / 5

v - 0 = 9.8x5

v = 49 m/s

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AlexFokin [52]

Answer:

$8.67791\times 10^{25}\ kg$

$0.34589\ m/s^2$

$0.07903\ m/s^2$

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M = Mass of Uranus

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Uranus = 25360 km

h = Altitude = 104000 km

$r_m$ = Radius of Miranda = 236 km

m = Mass of Miranda = $6.6\times 10^{19}\ kg$

Acceleration due to gravity is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow M=\dfrac{gr^2}{G}\\\Rightarrow M=\dfrac{9\times 25360000^2}{6.67\times 10^{-11}}\\\Rightarrow M=8.67791\times 10^{25}\ kg$

The mass of Uranus is $8.67791\times 10^{25}\ kg$

Acceleration is given by

$a_m=\dfrac{GM}{(r+h)^2}\\\Rightarrow a_m=\dfrac{6.67\times 10^{-11}\times 8.67791\times 10^{25}}{(25360000+104000000)^2}\\\Rightarrow a_m=0.34589\ m/s^2$

Miranda's acceleration due to its orbital motion about Uranus is $0.34589\ m/s^2$

On Miranda

$g_m=\dfrac{Gm}{r_m^2}\\\Rightarrow g_m=\dfrac{6.67\times 10^{-11}\times 6.6\times 10^{19}}{236000^2}\\\Rightarrow g_m=0.07903\ m/s^2$

Acceleration due to Miranda's gravity at the surface of Miranda is $0.07903\ m/s^2$

No, both the objects will fall towards Uranus. Also, they are not stationary.

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