Ask question

Ask question

All categories

2 years ago

6

a rural mail carrier leaves the post office and drives 22.0 km in a Northely direction. then drives in a 60° south of east direc

tion for 47 km. what is the displacement from the post office?

1 answer:

Scrat [10]2 years ago

7 0

Answer:

<em>Good Luck!</em>

<em> 5 Stars</em>

<em />

Explanation:

You might be interested in

A wrench is accidentally dropped from the top of a mast on a sailboat. Will the wrench hit at the same place on the deck whether

alekssr [168]

Answer:

Yes, because the wrench is moving at the same speed as the sailboat.

The main difference is that a person on the ground would see the wrench moving diagonally, while a person on the boat would see the wrench falling straight down,

This difference in paths lead to the relativistic change in lengths.

8 0

3 years ago

After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48

masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

$15min*\frac{1hr}{60min}=0.25hr$

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

$V=\frac{distance}{time}$

the plane traveled a distance to the north of 48km so the velocity is:

$V=\frac{48km}{0.25hr}$

so

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

$V_{x}=42km/hr*cos(-19^{o} )=39.71 i$

and

$V_{y}=42km/hr*sin(-19^{o} )=-13.67 j$

So the velocity of the wind can be expressed as a vector as:

$V_{wind}=(39.71i - 13.67j)km/hr$

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

$V_{plane x}=V_{plane/wind x}+V_{wind x}$

$V_{plane/wind x}=V_{plane x}-V_{wind x}$

$V_{plane/wind x}=(0-39.71 i)km/hr$

$V_{plane/wind x}= -39.71 i km/hr$

and

$V_{plane y}=V_{plane/wind y}+V_{wind y}$

$V_{plane/wind y}=V_{plane y}-V_{wind y}$

$V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr$

$V_{plane/wind x}= 205.67 j km/hr$

So the velocity of the plane with respect to the wind can be rewritten as:

$V_{plane/wind x}= (-39.71i + 205.67 j) km/hr$

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

$speed=\sqrt{(-39.71)^{2}+(205.67)^{2} }$

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0

3 years ago

The magnitude of the magnetic field at a certain distance from a long, straight conductor is represented by B. What is the magni

masya89 [10]

Answer:

B/4

Explanation:

The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:

The field at twice the distance is B/4.

8 0

3 years ago

states that the force times the distance on one side of a lever's fulcrum equals the force times the distance on the other side

andrey2020 [161]

Answer:

Explanation:

It is said to be in equilibrium about the fulcrum . It is the principle of moments.

3 0

3 years ago

Physics Question - Matching

KiRa [710]

1) G

2) E

3) D

4) I

5) J

6) C

7) H

8) F

9) B

10) A

I think... i am not 100% sure....

7 0

3 years ago

Read 2 more answers