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aliya0001 [1]
1 year ago
10

Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to

trade. Astronaut 1 tosses the 0.130 kg apple toward astronaut 2 with a speed of vi,1 = 1.05 m/s . The 0.150 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.18 m/s . Unfortunately, the fruits collide, sending the orange off with a speed of 1.03 m/s in the negative y direction. What are the final speed and direction of the apple in this case?
Physics
1 answer:
german1 year ago
5 0

Answer: v= 1.23 m/s θ = 75.3º

Explanation:

First of all, we define the direction in which both fruits are tossed as the x axis, so all initial momenta have horizontal components only.

Now, if no external forces act during collision (due to the infinitesimal time during which collision takes place) momentum must be conserved.

As momentum is a vector, both components must be conserved, so we can write the following equations:

p₁ₓ = p₂ₓ ⇒ -m₁ . vi₁ +m₂. vi₂ = m₁ . vf₁ . cos θ  (1)

p₁y = p₂y ⇒ 0 =m₂ . vf₂ - m₁. vf₁. sin θ (2)

Replacing by the values of m1, m₂, vi₁, vi₂, and vf₂, we can calculate the value of the angle θ, that the apple forms with the horizontal, as follows:

(1) -0.13 Kg. 1.05 m/s + 0.15 Kg. 1.18 m/s = 0.13. vf . cos θ  

(2) 0.15 Kg. 1.03 m/s = 0.13 vf. sin θ

sin θ / cos θ = 3.82 ⇒ tg θ = 3.82 ⇒ θ = arc tg (3.82) = 75.3º

Replacing this value of θ in (2), we get:

0.15 kg. 1.03 m/s = 0.13 vf . sin 75.3º = 0.13 . vf . 0.967

Solving for vf, we get:

vf = 0.15 kg. 1.03 m/s / 0.13. 0.967 = 1.23 m/s

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Answer:

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Exlanation:

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solution

we get here binding energy using threshold frequency of the metal that is express as

E=h\nu_{o}    ..................1

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and here N is the Avogadro constant = 6.023\times10^{23}

so E will \frac{x}{6.023\times10^{23}}  

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x = 99770.99

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4 0
2 years ago
What is the absolute value of the horizontal force that each athlete exerts against the ground?
alexandr402 [8]
Refer to the diagram shown below.

When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.

At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.

Answer:  
The horizontal force is μW,
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Kai takes a hamburger off of the grill and puts a piece of cheese on it. Explain how and why the cheese melts when Kai puts it o
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1 year ago
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Explanation:

1) You need to choose your axis. In this case North is vertical and positive, South is vertical and negative, East is horizontal and positive, and West is horizontal and negative.

2) The vector with the two positive components is a vector in the first quadrant (North and East). That is what North of East 25° means.

3) Regarding the other options:

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v = s/t = 300/300 = 1m/s

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3 0
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