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Aloiza [94]
3 years ago
15

An astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity c

hanges, each taking place in a time interval 11.3 s . What are the average acceleration in each interval?
1.) At the beginning of the interval the astronaut is moving toward the right along the x-axis at 15.0 m/s, and at the end of the interval she is moving toward the right at 5.30 m/s.
a = ___________-??
2.) At the beginning she is moving toward the left at 5.30 m/s. and at the end she is moving toward the left at 15.0 m/s.
a =_________ ??
3.) At the beginning she is moving toward the right at 15.0 m/s, and at the end she is moving toward the left at 15.0 m/s.
a = _________??
Physics
1 answer:
stiv31 [10]3 years ago
5 0

Answer

given,

time interval = 11.3 s

a) initial velocity, vi = 15 m/s

   final velocity, v_f = -5.30 m/s

      a = \dfrac{v_f-v_i}{t}

      a = \dfrac{-5.30 -15}{11.3}

             a = -1.79 m/s²

   the direction is along left side

b)  initial velocity, vi = -5.30 m/s

   final velocity, v_f = -15 m/s

      a = \dfrac{v_f-v_i}{t}

      a = \dfrac{-15-(-5.30)}{11.3}

             a = -0.858 m/s²

   the direction is along left side

c) initial velocity, vi = 15 m/s

   final velocity, v_f = -15 m/s

      a = \dfrac{v_f-v_i}{t}

      a = \dfrac{-15-(15)}{11.3}

             a = -2.65 m/s²

   the direction is along left side

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Answer:

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(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

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(b)the electrical power generated for a breezy winter day

T_s = T_{\infty} = -15 ^oC = 258 \ k

[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \  \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k

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