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IrinaK [193]
2 years ago
15

A 2.50-kg projectile is launched at an angle of 20.0above the horizontal with an initial speed of 100m / s from the top of a bui

lding 40 m high. If the projectile is traveling 85.0 m/ s at its maximum height of 47.0 m, how much work has been done on the projectile by air friction ?
Physics
1 answer:
Reptile [31]2 years ago
7 0

The amount of work that has been done on the projectile by air friction is 7,569.2 J.

<h3>Work done by air friction on the projectile</h3>

The work done by air friction on the projectile is calculated as follows;

W = ΔK.E

W = ¹/₂mvf² - ¹/₂mv₀²

where;

  • vf is the final velocity = 85 m/s
  • v₀ is the initial vertical velocity = 100 x sin(20) = 34.2 m/s

W = ¹/₂m(vf² - v₀²)

W = ¹/₂ x 2.5 (85² - 34.2²)

W =  7,569.2 J

Thus, the amount of work that has been done on the projectile by air friction is 7,569.2 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

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the half-life of carbon-14 is 5,730 years. After 11,460 year, how much of original carbon-14 remains?
Inessa [10]

Via the half-life equation:

A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}

Where the time elapse is 11,460 year and the half-life is 5,730 years.

A_{final}=A_{initial}(\frac{1}{2})^\frac{11460}{5730} \\\\A_{final}=A_{initial}\frac{1}{4} \\\\A_{final}=\frac{1}{4}A_{initial}

Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.

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Plants absorb water from the soil through their roots. _______ of this water is used for photosynthesis; _______ of this water e
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THE CORRECT CHOICE IS B
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3 years ago
Read 2 more answers
IV) Fill in the blanks:
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6.)"Conduction" is the primary mode of heat transfer in liquid and gases.

I hope this helps you...

5 0
3 years ago
Arrange the examples in order, starting with the object that has the least amount of energy. In each case, assume there’s no fri
Artemon [7]
First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,

Ep=m*g*h=0.75*9.8*1.5=11.025 J

Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,

E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J

Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek

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Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,

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The order of examples starting with the lowest energy:

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4 0
3 years ago
How could you test the hypothesis that elephants interpreted the ground signal as being farther away than the air signal?
bekas [8.4K]

Answer:

One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.

Explanation:

This hypothesis is based on the fact that the speed of sound in air is v = 343 m / s with a small variation with temperature.

The speed of sound in solid soil is an average of the speed of its constituent media, giving values ​​between

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 concrete 4000 m / s

 fabrics     1540 m / s

 earth       5000 m / s wave S

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we can see that the speed on solid earth is an order of magnitude greater than in air.

One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.

From the initial information, the wave going through the ground should arrive first.

3 0
4 years ago
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