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natulia [17]
4 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]4 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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Help please! Will give brainly! and thanks.
kipiarov [429]
<h2>Answer:</h2>

All the energy sources are correctly matched with their category.

Explanation:

Renewable energy sources:

These are energy sources which can be replenished as they don't in involves the irreversible phase change.

These resources can never be ended as they can be used again and again.

Wind, geothermal, biomass, bio gas are example of renewable energy sources.

Non renewable energy sources:

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7 0
4 years ago
Help for brainlist easy science plz
zalisa [80]

Answer:

the pics upside down fam

Explanation:

6 0
3 years ago
Read 2 more answers
B c<br>_ - _<br>a b<br>( a=3 b=-5 c=6)​
Pachacha [2.7K]
(-5)/3 - 6/(-5)
You can solve it now :)
7 0
3 years ago
What is the acceleration of the object?<br> m/s²
Orlov [11]

Answer:

-2.5m/s^2

Explanation:

10-40/12-0=-2.5

7 0
2 years ago
Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This
vova2212 [387]

Answer:

1.554\times 10^{32}\ \text{kg}

Explanation:

M = Mass of each star

T = Time period = 15.5 days

v = Orbital velocity = 230 km/s

G = Gravitational constant = 6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2

Radius of orbit is given by

R=\dfrac{vT}{2\pi}

We have the relation

\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}

The mass of each star is 1.554\times 10^{32}\ \text{kg}

6 0
3 years ago
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