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natulia [17]
3 years ago
12

When a jet lands on an aircraft carrier, a hook on the tail of the plane grabs a wire that quickly brings the plane to a halt be

fore it overshoots the deck. in a typical landing, a jet touching down at 240 km/h is stopped in a distance of 95 m .\?
Physics
1 answer:
Veronika [31]3 years ago
8 0
The problem seems to be incomplete because there is no question. However, from the problem description, the logical question is to find he acceleration needed by the jet to land on the airplane carrier. The working equation would be:

2ad = v₂² - v₁²
Since the jet stops, v₂ = 0. Substituting the values:
2(a)(95 m) = 0² - [(240 km/h)(1000 m/1 km)(1h/3600 s)]²
Solving for a,
<em>a = -23.39 m/s² (the negative sign indicates that the jet is decelerating)</em>
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A balloon will stick to a wooden wall if the balloon is charged
grigory [225]
<h2>Answer: either way</h2>

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Then, when two objects come into contact, the electrons of one of them can become part of the other.

Thus, by bringing the balloon closer to the wall, the wall, which is also made up of atoms, will reorder its charges in such a way that its electrons or protons become part of the balloon, charging it.

7 0
2 years ago
2. What is the water pressure at a depth of 24 m in a lake? [Density of water, p = 1 000 kg m- and gravitational acceleration, g
Andrei [34K]

Answer:

Pressure = ρgh

pressure (p) is measured in pascals (Pa)

density (ρ) is measured in kilograms per metre cubed (kg/m3)

The fore of gravitational field strength (g) is measured in N/kg or m/s 2

height of column (h) is measured in metres (m)

Answer = 235,200 Pa

Explanation:

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4 0
2 years ago
17. A volleyball weighs about 300 grams.
atroni [7]

Answer:

PE = 44.1 J

Explanation:

Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)

We have:

  • 1 kilograms = 1000 grams.

We convert it using a rule of 3, replacing, simplifying units and solving:

  • \boxed{\bold{x=\frac{gr*1\ kg}{1000\ gr}=\frac{300\ gr*1\ kg}{1000\ gr}=\frac{300\ kg}{1000}=\boxed{\bold{0.3\ kg}}}}

==================================================================

Earth's gravity is known to be 9.8 m/s², so we have:

Data:

  • m = 0.3 kg
  • g = 9.8 m/s²
  • h = 15 m
  • PE = ?

Use formula of potencial energy:

  • \boxed{\bold{PE=m*g*h}}

Replace and solve:

  • \boxed{\bold{PE=0.3\ kg*9.8\frac{m}{s^{2}}*15\ m}}
  • \boxed{\boxed{\bold{PE=44.1\ J}}}

Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.

The potential energy of the volleyball is <u>44.1 Joules.</u>

Greetings.

8 0
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The answer is D-Testable

Hope this helps
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