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3 years ago

The Hurricane's "eye" is surrounded by a "wall" of fast moving air? True or False

2 answers:

7 0

True because The Eye Wall of a hurricane's most devastating region. Located just outside of the eye is the eye wall. This is the location within a hurricane where the most damaging winds and intense rainfall is found. At the surface, the winds are rushing towards the center of a hurricane forcing air upwards at the center.

natulia [17]3 years ago

3 0

Answer:TRUE

Explanation:

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The heat capacity of nickel is 0.444 J/(g · °C). Calculate the amount of heat needed to raise the temperature of 18 g of nickel

azamat

The final speed of the nickel at the given quantity of heat is determined as 202.1 m/s.

<h3>Final speed of the nickel</h3>

Apply the principle of conservation of energy.

Q = mcΔθ

Q = (18)(0.444)(66 - 20)

Q = 367.63 J

Q = K.E = ¹/₂mv²

2K.E = mv²

v = √(2K.E/m)

where;

v is the final speed

v = √(2 x 367.63)/(0.018))

v = 202.1 m/s

Learn more about speed here: brainly.com/question/4931057

#SPJ1

5 0

1 year ago

An arrow is moving at 35 m/s and travels for 5 seconds. how far did the arrow travel?

pashok25 [27]

The answer is 175 meters

8 0

3 years ago

Describe a situation where you can be traveling at a low speed but have an extremely high velocity

Reil [10]

Answer:

Cruising at 35,000 feet in an airliner, straight toward the east,

at 500 miles per hour

Explanation:

3 0

2 years ago

Read 2 more answers

If a person is walking at 1.2 m/s and 60 seconds later the person is running at 10 m/s, what was the acceleration rate?

Marina CMI [18]

The acceleration rate would be .14667 m/s^2

6 0

3 years ago

A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$

Finally, using the Parallel Axis Theorem, we calculate I_B:

$I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}$

5 0

3 years ago