Ask question

Ask question

All categories

3 years ago

9

A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her orig

inal course, and flies 2 h in the new direction. If she maintains a constant speed of 690 mi/h, how far is she from her starting position

1 answer:

r-ruslan [8.4K]3 years ago

4 0

Answer:

2406 miles

Explanation:

Let A be the starting position, B the junction position and C the final position after flying the 3.5 hrs. Also, let b be the distance from the starting point:

$\angle ABC =(180-10) \textdegree=170\textdegree$

#Distance traveled in 1.5hrs is;

$c=690x1.5\\=1035mi$

#Distance traveled in next two hrs:

$a=690\times 2\\=1380mi$

#Now using the Cosine Rule:

$b^2=a^2+c^2-2ab\cos B\\\\=1380^2+1035^2-2(1380)(1035)cos170\textdegree\\\\b^2=5788.83\\\\b\approx 2406.00 \ mi$

Hence, the pilot is 2406 miles from her starting position.

You might be interested in

What is the driving force for (a) heat transfer, (b) electric current, and (c) fluid flow?

lara31 [8.8K]

Answer:

The driving force for (a) heat transfer is temperature difference. (b) electric current is voltage difference. (c) fluid flow is pressure or hydraulic head difference.

Explanation: (a) The driving force for heat transfer is temperature difference. Heat transfer between two mediums is possible only if the two mediums are at different temperature, the higher the temperature, the higher the heat transfer.

(b) The driving force for electric current is voltage difference. Voltage difference is defined as the potential difference in charge between two points in electrical field. For electric current to occur,the voltage must be high.

(c) The driving force for fluid flow is pressure difference or hydraulic head difference. For fluid to move upward,it requires energy.

3 0

3 years ago

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

3 years ago

Vesna [10]

Complete Question

A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward.

a

From this motion what is the distance covered

b

What is the magnitude and direction of the displacement

Answer:

a

$D = 72 \ m$

b

Magnitude

$d = - 20 \ m$

Direction

West

Explanation:

From the question we are told that

The first distance covered westward is $d_w_1 = 18 \ m /tex] The first distance covered eastward is [tex]d_e1 = 12 \ m /tex] The second distance covered westward is [tex]d_w_2 = 28 \ m /tex] The second distance covered eastward is [tex]d_e2 = 14 \ m /tex] Generally the distance covered is mathematically represented as [tex]D = d_w1 + d_w2 + d_e1 + d_e2$

=> $D = 18 + 28 + 12 + 14$

=> $D = 72 \ m$

For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative

The magnitude of the displacement is

$d = -d_w1 -d_w2 + d_e1 + d_e2$

=> $d = -18-28 + 12 + 14$

=> $d = - 20 \ m$

The direction is west

7 0

3 years ago

What do electric forces between charges depend on

denis23 [38]

Answer:

On the magnitude of the charges, on their separation and on the sign of the charges

Explanation:

The magnitude of the electric force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

From the formula, we see that the magnitude of the force depends on the following factors:

- magnitude of the two charges

- separation between the charges

Moreover, the direction of the force depends on the sign of the two charges. In fact:

- if the two charges have same sign, the force is repulsive

- if the two charges have opposite signs, the force is attractive

7 0

3 years ago

What is the momentum of an object with 5.22 kg and 1.7 m/s

BlackZzzverrR [31]

Answer:

8.874

Explanation:

You need to times 5.22 kg and 1.7 m/s to get 8.874.

8 0

3 years ago