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3 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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Answer

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3 years ago

Read 2 more answers

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

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3 years ago

The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,

Travka [436]

1) In a circular motion, the angular displacement $\theta$ is given by
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where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: $S=2600 m$ . Using the information about the radius, $r=0.35 m$ , we find the total angular displacement:
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2) If we put larger tires, with radius $r=0.60 m$ , the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
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