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cestrela7 [59]
2 years ago
5

Two astronauts are playing catch in a zero gravitational field. Astronaut 1 of mass m1 is initially moving to the right with spe

ed v1 . Astronaut 2 of mass m2 is initially moving to the right with speed v2>v1 . Astronaut 1 throws a ball of mass m with speed u relative to herself in a direction opposite to her motion. Astronaut 2 catches the ball. The final speed of astronaut 1 is vf,1 and the final speed of astronaut 2 is vf,2 .What is the speed vf,1 of astronaut 1 after throwing the ball?
Physics
1 answer:
Ede4ka [16]2 years ago
5 0

The final velocity (v_1_f) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut (v_2_f) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts <em>after throwing the ball</em>.

The given parameters;

  • Mass of the first astronaut, = m₁
  • Mass of the second astronaut, = m₂
  • Initial velocity of the first astronaut, = v₁
  • Initial velocity of the second astronaut, = v₂ > v₁
  • Mass of the ball, = m
  • Speed of the ball, = u
  • Final velocity of the first astronaut, = v_f_1
  • Final velocity of the second astronaut, = v_f_2

The final velocity of the first astronaut relative to the second astronaut after throwing the ball is determined by applying the principle of conservation of linear momentum.

m_1v_1 + m_2v_2 = m_2v_2_f + m_1v_1_f

if v₂ > v₁, then v_1_f > v_2_f, to conserve the linear momentum.

Thus, the final velocity (v_1_f) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut (v_2_f) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts after throwing the ball.

Learn more here: brainly.com/question/24424291

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Answer:

<em>The magnitude of vector d is 16 and the angle with the x-axis is 270°</em>

Explanation:

<u>Operations With Vectors</u>

Given two vectors in rectangular components:

\vec a=(ax,ay)\ ,\  \vec b=(bx,by)

The sum of the vectors is:

\vec a+\vec b=(ax+bx,ay+by)

The difference between the vectors is:

\vec a-\vec b=(ax-bx,ay-by)

The magnitude of \vec a is:

|\vec a|=\sqrt{ax^2+ay^2}

The angle \vec a makes with the horizontal positive direction is:

\displaystyle \tan\theta=\frac{ay}{ax}\\

The question provides the vectors:

\vec a=(2,6)

\vec b=(2,22)

\vec d=\vec a-\vec b

Calculate:

\vec d=(2,6)-(2,22)=(0,-16)

The magnitude of \vec d is:

|\vec d|=\sqrt{0^2+(-16)^2}=\sqrt{0+256}=16

The angle is calculated by:

\displaystyle \tan\theta=\frac{-16}{0}

The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.

The magnitude of vector d is 16 and the angle with the x-axis is 270°

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3.46 %

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The maximum possible efficiency is given by \eta =1-\frac{T_c}{T_h}

So maximum efficiency \eta =1-\frac{T_c}{T_h}=1-\frac{279}{289}=0.0346=3.46 %

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Answer:

1. balanced

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You have it exactly backwards.

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