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Sloan [31]
3 years ago
8

To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeof

f velocity, depends on several factors, including the weight of the aircraft and the wind velocity.
Physics
1 answer:
il63 [147K]3 years ago
6 0

<h2> The potential and kinetic energy of airplane are affected by these factors </h2>

Explanation:

When airplane rises up , it requires potential energy . This potential energy can be taken from the kinetic energy of airplane .

Thus if the speed of wind is larger , it can either oppose the motion of velocity or can favour the velocity of airplane  . By which its kinetic energy is effected .

If the weight of airplane is changed , it will effect the potential energy required . Thus heavier plane requires higher potential energy for attaining the same height .

Thus these two factor has important role in the flight of airplane .

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Yes it is work because when you throw a ball, you transfer energy to it and it moves.
7 0
2 years ago
What is the approximate terminal velocity of a sky diver before the parachute opens
anzhelika [568]

Answer:

The approximate terminal velocity of a sky diver before the parachute opens is 320 km/h.

Explanation:

  • The terminal velocity is the maximum magnitude of velocity that is attained by the diver when he or she falls in the air.
  • The terminal velocity of the person diving in air before opening parachute is 320 km/h that means the velocity when the person is experiencing free fall is 320 km/h.
  • During terminal velocity, we can represent mathematical equation as;

                           Buoyancy force + drag force = Gravity

6 0
3 years ago
Select the correct statement to describe when a sample of liquid water vaporizes into water vapor. Question 12 options: Temperat
pashok25 [27]
<span>temperature increases and molecular motion increases while shape becomes less defined.


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3 0
3 years ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
3 years ago
Use the general formulas for gravitational force and centripetal force to derive the relationship between speed (v) and orbital
german

Solution :

We know that :

Formula for Gravitational force is given by :

$F_g=\frac{Gmn}{r^2}$

where, G is the gravitational constant

            M is the mass of the bigger body

            m is the mass of the smaller body

            r  is the distance between the two bodies.

And the formula for the centripetal force is given by :

$F_c=\frac{mv^2}{r}$

where, m is the mass of the rotating body

            v is the velocity

             r is the radius of rotation of the body.

We know that mathematically, the gravitational force is equal to the centripetal force of the body.

Therefore,

$F_g=F_c$

$\frac{GMm}{r^2}=\frac{mv^2}{r}$

$\sqrt{\frac{GM}{r}}=v$

Hence derived.

5 0
2 years ago
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