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Gnesinka [82]
1 year ago
6

A recreational flyer does not need an airspace authorization in which type of airspace?.

Physics
1 answer:
finlep [7]1 year ago
5 0

A recreational flyer does not need an airspace authorization in uncontrolled airspace.

An airspace in which the Air Traffic Control service is not considered necessary is called uncontrolled airspace.

In general, an uncontrolled airspace remains at 400 or under 400 levels surpassing the ground.

Moreover, in an uncontrolled airspace the recreational flyer does not need any prior authorization and should also have particular knowledge about the airport traffic patterns

To put it simply, the remote pilots and flyers are not allowed to use landing and take off areas in an uncontrolled airspace.

If you need to learn more about airspace click here:

brainly.com/question/14451166

#SPJ4

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James and John dive from an overhang into the lake below. James simply drops straight down from the edge. John takes a running s
liraira [26]

Answer:

Both of them reach the lake at the same time.

Explanation:

We have equation of motion s = ut + 0.5at²

Vertical motion of James : -

          Initial velocity, u = 0 m/s

         Acceleration, a = g

         Displacement, s = h

    Substituting,

                  s = ut + 0.5 at²

                 h = 0 x t + 0.5 x g x t²

                 t_{James}=\sqrt{\frac{2h}{g}}

Vertical motion of John : -

          Initial velocity, u = 0 m/s

         Acceleration, a = g

         Displacement, s = h

    Substituting,

                  s = ut + 0.5 at²

                 h = 0 x t + 0.5 x g x t²

                 t_{John}=\sqrt{\frac{2h}{g}}

So both times are same.

Both of them reach the lake at the same time.

3 0
3 years ago
Which of the following objects has more kinetic energy than it does potential energy? Select the correct answer below:______1.a
Lilit [14]

Answer:

The pendulum of the clock.

Explanation:

Hi there!

The kinetic energy is the energy associated with the velocity of the object. The potential energy is the energy associated with the position of the object. In the objects listed in the question, only one object is moving: the pendulum of the clock (assuming that the clock is functioning). If the clock functions, the pendulum is moving when it is at the lowest point of its arc of motion and with maximum velocity. All potential energy that the pendulum stored when it reached the highest height, is transformed into kinetic energy at the lowest point. Thus, at that point, the object has more kinetic energy than potential energy.

6 0
3 years ago
You are attempting to row across a stream in your rowboat. Your paddling speed relative to still water is 3.0 m/s (i.e., if you
Nataliya [291]

Answer:

Please check the attached file for the diagram

Explanation:

The velocity of the of the rowboat V_{tot}  through the river is the resultant velocity. It is obtained taking a vector sum of the velocity in still water and the velocity of the river.

There are several ways to take this vector sum, but the question makes it simple for us to use Pythagoras's theorem because the East and North directions are perpendicular to each other.

Hence;

V_{tot}^2=V_{still}^2+V_{w}^2\\V_{tot}^2=3^2+4^2

V_{tot}=\sqrt{3^2+4^2}\\ V_{tot}=\sqrt{25}=5m/s

6 0
3 years ago
The maximum wavelength an electromagnetic wave can have and still eject an electron from a copper surface is 264 nm .What is the
Tamiku [17]

Answer:

4.71 eV

Explanation:

For an electromagnetic wave with wavelength

\lambda=264 nm = 2.64\cdot 10^{-7} m

the energy of the photons in the wave is given by

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2.64\cdot 10^{-7}m}=7.53\cdot 10^{-19} J

where h is the Planck constant and c the speed of light. Therefore, this is the minimum energy that a photon should have in order to extract a photoelectron from the copper surface.

The work function of a metal is the minimum energy required by the incident light in order to extract photoelectrons from the metal's surface. Therefore, the work function corresponds to the energy we found previously. By converting it into electronvolts, we find:

E=\frac{7.53\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=4.71 eV

3 0
2 years ago
Do you think there is water on exoplanets
Ksivusya [100]

Answer:

Explanation:

Some exoplanets may depending on the climate and vicinity from the sun.

8 0
3 years ago
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