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2 years ago

I am starting out as a pilot. What single engine make of airplane should I get (Cheaper to bye one and train than to rent it)?

2 answers:

7 0

You have to make sure you know what your doing plus if you have your license and stuff to fly you have to practice on a 25 minute course

DiKsa [7]2 years ago

4 0

You can't be serious ! You might consider the Cessna 172.

You might be interested in

Which colors of light represents the lowest visible frequency?

kodGreya [7K]

Answer:

Red

Explanation:

Red is a colour which has the lowest frequency. Violet has the highest frequency. Frequency has a direct relationship with energy. This means the higher the frequency, the higher the energy. Red has the lowest energy of all the colors too.

The frequency and Energy has an inverse relationship with the wavelength.

However Red has the longest wavelength of about 620 - 780 nanometer.

3 0

3 years ago

Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is

attashe74 [19]

Answer:

0.125 volts

Explanation:

The induced emf can be sufficient to stimulate neuronal activity.

One such device generates a magnetic field within the brain that rises from zero to 1.5 T in 120 ms.

We need to find the induced emf within a circle of tissue of radius 1.6 mm and that is perpendicular to the direction of the field. The formula for the induced emf is given by :

$\epsilon=-\dfrac{d\phi}{dt}$

Where

$\phi$ is magnetic flux

So,

$\epsilon=-\dfrac{d(BA)}{dt}\\\\=2\pi r\times \dfrac{dB}{dt}\\\\=2\pi \times 1.6\times 10^{-3}\times \dfrac{1.5-0}{120\times 10^{-3}}\\\\=0.125\ V$

So, the induced emf is equal to 0.125 volts.

7 0

3 years ago

Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the

Stells [14]

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0

3 years ago

Hydrogen gas is generated when acids come into contact with certain metals. When excess hydrochloric acid reacts with 2.2 g of (

CaHeK987 [17]

To solve the problem it is necessary to apply the concepts related to Byle's Law and Avogadro's Law.

The ideal gas equation would help us find the final solution to the problem, defined by

$PV = nRT$

Where,

T= Temperature of the gas

R = Universal as constant

n = number of moles

V = Volume

P = Pressure

For our case we have that the mass of Zn is 2.2g in moles would be

$[tex]Zn = \frac{2.2}{65}$ [/tex]

$Zn = 0.0338$

We know that 1 mole of hydrogen gas is proceed by 1 mole of zinc and the result is $Zn^{2+}$ , then Hydrogen can produce the same quantity,

$H_2 = 0.0338$

Applying the previous equation we have that

$V= \frac{nRT}{P}$

$V = \frac{0.0338*0.08206*293.15}{0.98}$

$V = 0.829L$

Therefore the volume of hydrogen gas is collected is 0.829L

6 0

3 years ago

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun

AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561 × 10¹⁹ electrons per second = 1.561 × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

8 0

3 years ago