Ask question

Ask question

All categories

3 years ago

12

you're reading from the journal of a European explorer from the early 1600s. In one passage, the explorer describes itting on th

e Atlantic Ocean with little wind. He describes the area as being quite far north of the equator and having nigh pressure. n which type of global wind was this explorer sailing? doldrums horse latitudes jet stream O polar easterlies

2 answers:

Zepler [3.9K]3 years ago

8 0

Answer: horse latitudes

Explanation:

Karo-lina-s [1.5K]3 years ago

6 0

Answer:

B

Explanation:

Doldrums are located some 5 degrees away from north while horse latitude is located some thirty degrees away from north latitude and by same degrees from the southern hemisphere. Polar eastariles are cold winds blowing across the Polar Regions and jet stream is fast blowing winds.

Horse latitude is silent winds and the region has high pressure and very little precipitation.

Hence, option B is correct

You might be interested in

Which of the following best describes a plane?

zloy xaker [14]

On that list of choices, 'C' is the only "example" of a plane.
None of the choices "describes" a plane.

8 0

3 years ago

A pencil rolls horizontally of a 1 meter high desk and lands .25 meters from the base of the desk. How fast was the pencil rolli

vovikov84 [41]

Answer: 0.55 m/s

Explanation:

This situation is related to projectile motion (also called parabolic motion), where the main equations are as follows:

$x=V_{o} cos\theta t$ (1)

$y=y_{o}+Vo sin \theta t + \frac{g}{2}t^{2}$ (2)

Where:

$x=0.25 m$ is the horizontal displacement of the pencil

$V_{o}$ is the pencil's initial velocity

$\theta=0\°$ since we are told the pencil rolls <u>horizontally</u> before falling

$t$ is the time since the pencil falls until it hits the ground

$y_{o}=1 m$ is the initial height of the pencil

$y=0$ is the final height of the pencil (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity, always acting vertically downwards

Begining with (1):

$x=V_{o} cos(0\°) t$ (3)

$x=V_{o}t$ (4)

Finding $t$ from (2):

$0=1 m+ \frac{-9.8m/s^{2}}{2}t^{2}$ (5)

$t=\sqrt{\frac{-2y_{o}}{g}}$ (6)

Substituting (6) in (4):

$x=V_{o}\sqrt{\frac{-2y_{o}}{g}}$ (7)

Isolating $V_{o}$ :

$V_{o}=\frac{x}{\sqrt{\frac{-2y_{o}}{g}}}$ (8)

$V_{o}=\frac{0.25 m}{\sqrt{\frac{-2(1 m)}{-9.8m/s^{2}}}}$ (9)

Finally:

$V_{o}=0.55 m/s$

4 0

4 years ago

Multiply the following numbers, using scientific notation and the correct amount of significant digits. 1.003 m⋅3.09 = _____ 3.0

lozanna [386]

Answer

correct answer is 3.10

Explanation:

in this question we have to multiply two numbers 1.003 and 3.09.

1.003 has 4 significant digits and 3.09 has 3 significant digits so answer must have 3 significant digits.

$1.003\times 3.09=3.09927$

$1.003\times 3.09=3.10$

Hope it will help you

3 0

3 years ago

Read 2 more answers

Match the choices to the appropriate blank . Use each choice only once.

pav-90 [236]

Answer:

1. about 1.5 AU

2. about 5 AU

3. about 8 light-years

4. about 100,000 light-years

5. less than 0.01 AU

Explanation:

a. Mars is about 1.5 AU from the Sun.

b. Jupiter is about 5 AU from the Sun.

c. The star Sirius is about 8 light-years from the Sun.

d. The diameter of the Milky Way Galaxy is about 100,000 light-years.

e. The distance from Earth to the Moon is less than 0.01 AU.

Note: AU is an acronym for Astronomical Unit and it is a standard unit by astronomers to illustrate the distance between the planetary bodies found in the solar system.

5 0

3 years ago

A 1250 kg car is pulling a 325 kb trailer. The car and trailer have an acceleration of 2,15m/sec2 together. Determine the net fo

xenn [34]

To solve this problem we will apply the definition of Newton's second law, which says that force is equivalent to body mass by acceleration. In this case the mass of the trailer is 325Kg and its acceleration is $2.15m / s ^ 2$ , so we will proceed to replace and multiply these values to find the net force on this object.

$F= ma$

$F= (325 kg)(2.15 m/s^2)$

F= 698.75N

Under the reference system in which the direction of travel is the positive direction, the direction of the force will be positive.

7 0

3 years ago