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photoshop1234 [79]
3 years ago
7

How will frost on the wings of an airplane affect takeoff performance?

Physics
1 answer:
san4es73 [151]3 years ago
8 0

Frost will disturb the smooth flow of air over the wing, unpleasantly distressing its lifting competence. In other words, this spoils the even flow of air over the wings, by this means decreasing lifting capability. Also, frost may avoid the airplane from becoming flying at normal departure speed.

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Infer whether a system can have kinetic energy and potential energy at the same time
Rasek [7]
Yes, an object can have both of these at the same time. Potential energy is energy that is stored in an object. Kinetic energy is the energy that is associated with motion. So what you have to have is an object that is in motion but still has more energy that it has yet to convert into kinetic energy.
4 0
2 years ago
Which of the objects below has the greatest acceleration? *
lesya692 [45]

\\ \bull\sf\dashrightarrow F=ma

\\ \bull\sf\dashrightarrow a=\dfrac{F}{m}

Now

\\ \bull\sf\dashrightarrow a\propto\dfrac{1}{m}

\\ \bull\sf\dashrightarrow a\propto F

  • Lower mass=Higher acceleration
  • Lower Force=Lower Acceleration

Option B has lowest mass and highest force hence its correct

8 0
2 years ago
HELP ASAP!<br> Everything on screenshot!
Oksana_A [137]
This is a divergent plate boundary because the two tectonic plates are colliding together.
3 0
3 years ago
Supposing d(t) is known to have value D,
creativ13 [48]

Answer:

  • The procedure is: solve the quadratic equation for t.

Explanation:

This question assumes uniformly accelerated motion, for which the distance d a particle travels in time t is given by the general equation:

  • d(t)=d_0+v_0t+at^2/2

That is a quadratic equation, where the independent variable is the time t.

Thus, the procedure that will find the time t at which the distance value is known to be D is to solve the quadratic equation for t.

To solve it you start by changing the equation to the general form of the quadratic equations, rearranging the terms:

  • (a/2)t^2+v_0t+(d_0-D)=0

Some times that equation may be solved by factoring, and always it can be solved by using the quadratic formula:

  • t=\frac{-b+/-\sqrt{b^2-4ac} }{2a}

Where:

a=-a/2\\ \\ b=v_0\\ \\ c=d_0-D

That may have two solutions. Some times one of the solution makes no physical sense (for example time cannot be negative) but others the two solutions are valid.

5 0
3 years ago
Must new substance be formed when you observe property
d1i1m1o1n [39]
Of course not !
You can observe the color of a rose, measure the length of
your house, take the temperature of your hot chocolate, and
measure your own weight, without creating any new substances.
6 0
3 years ago
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