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2 years ago

What accounts for the two precipitation peaks in mbandaka?

1 answer:

7 0

The two precipitation peaks in Mbandaka during March to April and September to November is due to the intertropical convergence zone.

Intertropical convergence zone is a narrow zone located near the equator. It is where the northern and southern air masses intersect which results to low atmospheric pressure. Due to the intertropical convergence zone’s meeting of air masses, often times the air pressure are lower which will results to colder air, or even rainfall during the period of March to April, and most especially September to November in Mbandaka.

<span>Since Mbandaka is located at the cented of Tumba-Ngiri-Maindombe area, which is named as a Wetland of International importance, there is really a bigger chance that this place experience above 60mm precipitation in a year, temperatures averaging from 23 – 26 degrees Celsius.</span>

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B. A 20kg wagon is pulled up a 30° incline at an acceleration of 2.5ms?. The force pulling the wagon is parallel to the incline

yawa3891 [41]

The value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.

<h3>Coefficient of the kinetic friction</h3>

The value of coefficient of kinetic friction is calculated as follows;

F - Ff = ma

F - μmgcosθ = ma

where;

F is applied force
μ is coefficient of kinetic friction
m is mass of the wagon
a is acceleration of the wagon

182 - μ(20 x 9.8 x cos30) = 20(2.5)

182 - 169.74μ = 50

182 - 50 = 169.74μ

132 = 169.74μ

μ = 132/169.74

μ = 0.78

Thus, the value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.

Learn more about coefficient of friction here: brainly.com/question/20241845

7 0

2 years ago

A solid cylinder with a mass of 2.72 kg and a radius of 0.083 m starts from rest at a height of 4.20 m and rolls down a 88.7 ◦ s

Bingel [31]

Explanation:

According to the law of conservation of energy ,

Potential energy = kinetic energy

$mgh = \frac{1}{2} \times mv^{2} + \frac{1}{2} \times I \times \omega^{2}$

I = $\frac{mr^{2}}{2}$

$\omega = \frac{v}{r}$

$mgh = [\frac{1}{2} \times mv^{2}] + [\frac{1}{2} \times (\frac{mr^{2}}{2}) \frac{v^{2}}{r^{2}}]$

mgh = $[\frac{1}{2} \times mv^{2}] + [\frac{1}{4} \times mv^{2}]$

$g \times h = \frac{3}{4} \times v^{2}$

$9.8 \times 4.2 = \frac{3}{4} \times v^{2}$

v = 7.4 m/s

thus, we can conclude that the translational speed of the cylinder when it leaves the incline is 7.4 m/s.

5 0

2 years ago

Imagine you have a collection of identical flat-bottomed coffee filters that can be nested (stacked inside of each other) so tha

skad [1K]

Answer:

the terminal velocity of 14 nested coffee filters is 3.2 m/s

Explanation:

Given the data in the question;

we know that;

The terminal velocity is proportional to the square root of weight.

v ∝ √W

v = k√W

the proportionality constant depends upon the surface area and the density of the medium (like air). The coffee filters can be stacked such that the resulting area is roughly unchanged. So, the constant of proportionality k is also unchanged

v/√W = constant

v₂/√W₂ = v₁/√W₁

v₂ = v₁√(W₂ / W₁ )

given that;

v₁ = 0.856 m/s,

W₂ = 14W₁; meaning 14 coffee filters have 14 times the weight of a single coffee filter

so we substitute

v₂ = 0.856 √(14W₁ / W₁ )

v₂ = 0.856 √( 14( W₁/W₁)

v₂ = 0.856 √( 14(1)

v₂ = 0.856 √( 14 )

v₂ = 0.856 × 3.741657

v₂ = 3.2 m/s

Therefore, the terminal velocity of 14 nested coffee filters is 3.2 m/s

6 0

2 years ago

What is in the boron family with 3 valence electrons

olganol [36]

Boron, Aluminum, Gallium, Indium, Thallium

7 0

2 years ago

Diagram of an atom with labels

lina2011 [118]

See this. I hope you find your answer

6 0

3 years ago