Ask question

Ask question

All categories

3 years ago

13

An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed o

f the plane (its speed in still air) is 320.0 km/h (about 200 mi/h), in which direction should the pilot head?

1 answer:

SashulF [63]3 years ago

8 0

Answer:

14.48° North of West.

Explanation:

The conditions mentioned in the question has been drawn in the figure below where,

The vector CA represents the velocity of airplane,
θ is the angle from west pointing the direction of airplane in which it is required to travel,
Vector AB is the direction of wind, and,
Vector CB is the resultant direction.

As ΔABC is a right angle triangle, for calculating the direction of airplane so that it can travel to west, we need to calculate the direction of vector CA which can be calculated with the help of sine theta.

So,

$Sin$ θ $= \frac{80}{320} = \frac{1}{4}$

Thus the value of θ = 14.48°

So, the pilot should head towards 14.48° North of West.

You might be interested in

I just need x isolated

mamaluj [8]

Answer:

$x=\frac{-y+\sqrt{y^2+4rt} }{2r}$

$x=\frac{-y-\sqrt{y^2+4rt} }{2r}$

Explanation:

$rx+y=\frac{t}{x}\\\\x(rx+y)=(\frac{t}{x})x\\\\rx^2+yx=t\\\\rx^2+yx-t=t-t\\\\rx^2+yx-t=0$

Solve using the quadratic formula.

$x=\frac{-y+\sqrt{y^2+4rt} }{2r}$

$x=\frac{-y-\sqrt{y^2+4rt} }{2r}$

7 0

3 years ago

Read 2 more answers

Conduction is the movement of heat energy because of

Ira Lisetskai [31]

Answer:

Conduction occurs when a substance is heated, particles will gain more energy, and vibrate more. These molecules then bump into nearby particles and transfer some of their energy to them. This then continues and passes the energy from the hot end down to the colder end of the substance.

Explanation:

pls make me brainliest

6 0

2 years ago

Read 2 more answers

During an eclipse the sun the earth and the moon act as a light source,an obstacle or screen=

Zigmanuir [339]

Answer:

Un eclipse solar se produce cuando la luna se interpone en el camino de la luz del sol y proyecta su sombra en la Tierra. Eso significa que durante el día, la luna se mueve por delante del sol y se pone oscuro. ... Este eclipse total se produce aproximadamente cada año y medio en algún lugar de la Tierra.

Explanation:

espero que te

haya

servido

6 0

3 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

a)

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

b)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

c)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

3 years ago

Which form of electromagnetic radiation with a relatively high amount of energy is used to sanitize lab goggles?

Nostrana [21]

Answer:

sorry

Explanation:

i don't know this but when i find the answer i will type here

5 0

3 years ago