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3 years ago

5

Is the tropic of cancer north or south of the equator?

2 answers:

jonny [76]3 years ago

6 0

Answer:

The tropic of cancer is in the north.

Explanation:

This tropic is also known as the <u>northern tropic. </u> Its exact position is 23°26'12.2" north of the Equator.

It passes through the following countries:

Mexico

Algeria

Niger

Libya

Egypt

Saudi Arabia

India

Bangladesh

China

Myanmar

Taiwan

And also it passes through the oceans:

Pacific Ocean and Atlantic Ocean.

Daniel [21]3 years ago

3 0

It’s North of the equator

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bacteria and other prokaryotes are found deep underground in water and in the air. one micron 10^-6 is a typical length scale as

tankabanditka [31]

<span>I don't think there is enough information from the question. If you are asked to calculate the number of microbes in a certain area., then it should be that the total area or maybe the volume of that should be given so you can estimate the number of microbes. Hope this helps.</span>

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3 years ago

The flu is an infection caused by a pathogen. When a pathogen enters the body, which blood component helps us to attack the path

Arada [10]

Answer:

white blood cells

Explanation:

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3 years ago

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a centrifugal force ride, similar to the gravitron, spin at a speed of 22 revolutions per minute. if the diameter of the ride is

Margarita [4]

The linear speed of the passengers in kilometres/hour is 1.152 km/h.

The centrifugal force ride spins at a speed of 22 rev/min. The diameter of the ride is 14 meters.

We know,

Diameter = 2(Radius)

Radius = 14/2m

Radius = 7m

Radius of the centrifugal ride is 7m.

Also,

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22 rev = 44π radian

Angular Speed in rad/s = 44π/60

The relationship between the angular speed W and the linear speed V is given as,

V = W/R

R is the radius of the circle,

Now, putting the values,

V = 44π/(60×7)

V = 0.32 m/s.

So, the linear speed of the passengers is 0.32 m/s.

It is know, for conversion from m/s to km/h

1 m/s = 3600/1000 km/h

So, 0.32m/s in km/h is,

0.32m/s = 0.32 × 3600/1000

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9 months ago

D= ((vf+vi)/2) t <br> Solve for vi

erastovalidia [21]

= t(Vf + Vi/2)
<span>Vf + Vi/2 = d/t </span>
<span>Vf = (d/t) - Vi/2 </span>

<span>Answer: Vf = (d/t) - Vi/2 OR (2dt - Vit)/2t

</span>

8 0

3 years ago

Read 2 more answers

A 18 kg sled starts up a 28 degree incline with a speed of 2.3 m/s. The coefficient of kinetic friction between the sled and the

sergey [27]

Answer:

d = 0.391 m

Explanation:

given,

mass of sled = 18 kg

inclined at an angle of = 28°

kinetic friction between the sled and inclined = 0.25

using energy equation

$\dfrac{1}{2}mv^2- Fd - mgh = 0$

$\dfrac{1}{2}mv^2- (\mu mg cos\theta )d - mgdsin\theta = 0$

$\dfrac{1}{2}mv^2 = (\mu mg cos\theta )d + mgdsin\theta$

$\dfrac{1}{2}mv^2 = d mg (\mu cos\theta+sin\theta)$

$d = \dfrac{v^2}{2g (\mu cos\theta+sin\theta)}$

$d = \dfrac{2.3^2}{2\times 9.8 (0.25\times cos28^0+sin28^0)}$

d = 0.391 m

hence, the distance traveled by the sled is equal to 0.391 m

7 0

3 years ago