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Colt1911 [192]
3 years ago
7

Imagine you live next-door to your best friend. Your bedroom windows face each other, and you want to set up a way to communica

te privately using waves. You have these supplies:
a long string
a spring toy long enough to stretch between your windows
metal cans
a Morse code alphabet decoder (Morse code is a system that represents letters of the alphabet with dots and dashes) imagine you live next-door to your best friend
Physics
2 answers:
Alexxandr [17]3 years ago
8 0

Answer:

J3rk it in the window ;-)

Explanation:

alexdok [17]3 years ago
4 0

Answer:

A morse code alphabet decoder maybe?? I am confused

Explanation:

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Interactive Solution 6.39 presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a
mariarad [96]

(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

v_y^2 - u_y^2 = 2as

where, taking upward as positive direction:

v_y is the final vertical velocity

u_y = 0 is the initial vertical velocity (zero because the pebble is launched horizontally)

a=g=-9.8 m/s^2 is the acceleration of gravity

s = -25.0 m is the displacement

Solving for vy,

v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

v_x = 20.0 m/s

So, the final speed of the pebble as it strikes the ground is

v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

u_y = 20.0 m/s

So we can find the final vertical velocity using the same suvat equation as before:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

u_y = -20.0 m/s

Using again the same suvat equation:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

We notice that the final value of the speed is always the same in all the three parts, so it does not depend on the direction of launching. This is due to the law of conservation of energy: in fact, the initial mechanical energy of the pebble (kinetic+potential) is the same in all three cases (because the height h does not change, and the speed v does not change either), and the kinetic energy gained during the fall is also the same (since the pebble falls the same distance in all 3 cases), therefore the final speed must also be the same.

7 0
3 years ago
What’s the answer to this
ladessa [460]
Choices  1,  2,  and 4 . . . . . Yes

Choices  3  and 5 . . . . . No
6 0
3 years ago
If you know what a JW is LET ME KNOW and it is a type of ppl
lisabon 2012 [21]

Answer:

only thing I think of when I see that is 'Just Wondering'

Explanation:

3 0
2 years ago
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Pani-rosa [81]
The energy carried by one photon is directly proportional to its
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That's why when you get past visible light and on up through ultraviolet,
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The photon with the highest energy is a gamma-ray photon.


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zhenek [66]

Answer:

c. dioptre that's the answer.

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3 years ago
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