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trapecia [35]
3 years ago
13

The pilot of an airplane notes that the compass indicates heading west. The airplane's speed relative to the air is 100 km/h. Th

e air is moving in a wind at 25.0 km/h toward the north. Find the velocity, magnitude and direction of the airplane relative to the ground.
Physics
1 answer:
irinina [24]3 years ago
6 0
The airplane's speed relative to the ground is

                                     √ (100² + 25²)

                                 =  √ (10,000 + 625)

                                 =  √ 10,625

                                 =      103.08  km/hr .

The angle of its velocity north of west is
the angle whose tangent is  (25/100)

                         arctan(25/100) = 14° north of west .

                                             (bearing = 284°)
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<span>4 x 3.1416 x 1740² = ~38,046,032km² surface area.</span>
8 0
3 years ago
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A circular area with a radius of 6.50 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle du
White raven [17]

Answer:

a. \phi _B=3.0528\times10^-^3T\ m^2\\ \\b. \phi_B=1.83299\times10^-^3T\ m^2\\\\c.\phi_B=0

Explanation:

#Consider a circular area of radius R=2.98cm in the xy-plane at z=0. This means all the are vector points toward the +ve z-axis.

a. first, find the magnetic flux if the magnetic field has a magnitude of B=0.23T and points toward the +ve z-axis. The angle between the magnetic field and the area is \theta=0. Hence the magnetic flux:-

\phi _B=\int {\bar B} . d\bar A \\=\int BdAcos(\theta)=BAcos(0)=BA\\\\=\pi R^2B=\pi(6.50\times10^-^3m)^2(0.230T)\\=3.0528\times10^-^3 T\ m^2

Hence flux magnitude in +z direction is 3.0528\times10^-^3T \ m^2

b. We now find the magnetic flux when the field has a magnitude of <em>B=0.230T</em> and points at an angle of \theta=53.1\textdegree from the +z direction.

Magnetic flux is calculated as:

\phi _B=\int\bar B \bar dA\\=\int BdAcos (\theta)=BAcos(0)=BA\\=\pi R^2B=\pi(6.50\times 10^-^2m)^2(0.230T)\\=1.83299\times 10^-^3 T \ m^2

Hence the flux at an angle of 53.1\textdegree is 1.83299\times 10^-^3T \ m^2

c. We now need to find the magnetic flux if the field has a magnitue of B=0.230T and points in the direction of +y-direction. As with the previous parts, the magnetic flux will be calculated as:

\phi_B= \int\bar B \times d\bar A\\=\int BdAcos(\theta)\\=BAcos(90\textdegree)\\=0

Hence the magnetic flux in the +y-direction is zero.

3 0
3 years ago
PLEASE HELP!!
astraxan [27]

Answer:

Explanation:

\lambda\\ = v/f

^That is the formula we are going to use.

Now, we were given the speed (v), which is 20.

Now we need to find frequency, in order to solve for the wavelength.

Frequency is the amount of waves in a fixed unit of one second, meaning our F value is the value of 5 divided by 4.

5/4 = 1.25

Therefore our F is 1.25

Now lets plug it in

\lambda\\ = v/f

\lambda\\ = 20/1.25

\lambda\\ = 16

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