Hurricanes are powered by heat energy in the atmosphere and the ocean. Typically, one hurricane uses an amount of heat roughly equal to Z thousand atomic bombs. If the excess heat built up in Earth's atmosphere since 1850 went into producing hurricanes, about how many more hurricanes might we expect in the world every year?

Answers

Answer 1

Answer:

In an year about 6 major hurricanes and 3 minor storms occurs.

Explanation:

Hurricanes are increasing in number and frequency to the increasing impact of global warming. As current research states that climate change in the tropics results in an increased storm frequency. The presence of extra heat in the air or the oceans form the stoma that leads to the production of such energy. According to o the 200-2007 data shows in the huge increase in the storm frequency and this suggests the impact of climate warming.

Related Questions

Using the image, which area of the ocean floor would have the largest amount of pressure? Explain

Answers

Answer: the area were the most pressure is at the deep sea trench (the deepest part)

Explanation: because the more on top of you is the more pressure.

number of atoms of each element ammonium hydroxide?​

Answers

Answer:

Explanation:

Ammonium Hydroxide has the chemical formula of

NH4OH

There is 1 Nitrogen (N) = 1

There are 5 Hydrogens (H) = 5

There is 1 Oxygen  (O) = 1

In total, there are 7 atoms in 1 molecule of NH4OH

Jake appears to have several personalities. One is his dead cousin Bob, who
apparently bought a car Jake didn't know about. Jake said he had no memory
of buying the car. Jake also has a personality of a 12-year-old child. Jake
appears to be suffering from:
A. schizophrenia.
B. generalized anxiety disorder.
C. dissociative identity disorder.
D. major depression.
SUBMIT

Answers

The answer is C. Dissociative identity disorder

-

This should be easy to spot

Answer: DISSOCIATIVE IDENTITY DISORDER

Explanation:

Which of the following statements about Australian football is TRUE?
A.
The players tend to wear a large amount of padding.
B.
Goals are scored by kicking the ball through goalposts.
C.
Each team has a number of set plays for both offense and defense.
D.
The constant movement of the ball is similar to baseball.

Answers

Answer:

B.

Goals are scored by kicking the ball through goalposts.

Explanation:

8.
2
AI
6
HCI
AICI:
+
H



Answers

Answer:

I'm sorry I can't help you...

A5.0 kg TNT explosive, initially at rest, explodes into two pieces. One of the pieces weighing 2.0 kg flies off to

the left at 36

What is the final speed of the remaining piece?

Answers

Answer:

v = 24 m/s, rightwards

Explanation:

Given that,

The mass of TBT explosive = 5 kg

It explodes into two pieces.

One of the pieces weighing 2.0 kg flies off to  the left at 36 m/s. Let left be negative and right be positive.

The law of conservation of momentum holds here. Let v be the final speed of the remaining piece. So,

[tex]5\times 0=2\times (-36)+3\times v\\\\-72=-3v\\\\v=24\ m/s[/tex]

So, the final speed of the remaining piece is 24 m/s and it is in the right direction.

To Calculate Velocity and Average Velocity you can use the same formula as both.

True Or False?

Answers

Answer:

I believe it's True!    Brainliest??

Explanation: Hope you have a great day :)

A plane travels 4.0 km at an angle of 25◦ to the ground, then changes direction and travels 10 km at an angle of 16◦ to the ground. What is the magnitude of the plane's total displacement? Answer in units of km.

Answers

Answer:

6.1 km

Explanation:

Given that a plane travels 4.0 km at an angle of 25◦ to the ground, then changes direction and travels 10 km at an angle of 16◦ to the ground. What is the magnitude of the plane's total displacement? Answer in units of km

The magnitude of the total displacement D can be calculated by using cosine formula

Ø = 25 - 16 = 9 degree

D^2 = 4^2 + 10^2 - 2 × 4 × 10 × cos 9

D^2 = 16 + 100 - 80cos9

D^2 = 116 - 79.02

D = sqrt( 36.98)

D = 6.1 m

Therefore, the magnitude of the plane's total displacement is 6.1 km

What is the name of the force that opposes, or the opposite of gravity ???!!

Answers

Answer:

Tension.

tension is the name of force that opposes or goes opposite of gravity

Hope this helps!

Answer: Friction

Explanation: Friction opposes gravity and any motion in two surfaces.

Hope This Helped, Have A Great Day!

What work is the person doing doing to the box if the box is just being held and it is not moving?

Answers

Allowing for a platform in which the box holds potential energy

An alien from the newly discovers planet nine weighs himself on his planet and finds his weight to be 3200 N. When he stoves on earth he once again weighs himself and find his aight to be 800 N. What is the acceleration due to gravity on planet 9?

Answers

Answer:

2.45 m/s²

Explanation:

From the question,

On the Earth

W = mg.................. Equation 1

Where W = weight of the alien on the earth, m = mass of the alien on the earth, g = acceleration due to gravity of the earth.

Make m the subject of equation 1

m = W/g................... Equation 2

Given: W = 3200 N

Constant: 9.8 m/s²

Substitute these value into equation 2

m = 3200/9.8

m = 326.5 kg.

Similarly,

On planet 9,

W' = mg'............... Equation 3

Where W' = weight of the alien on planet 9, g' = acceleration due to gravity on planet 9.

make g' the subject of the equation

g' = W'/m............ Equation 4

Given: W' = 800 N

Substitute into equation 4

g' = 800/326.5

g' = 2.45 m/s²

which diagram represents the greatest magnitude of displacement for an object

Answers

Answer: if every arrow=1m do 1 times 1 times 1 times 1

Explanation:

(p.s. it equals 1 meter in area 4 meters in perimeter)

Calcular la densidad absoluta de un fluido cuya masa es de 12.8 kg y que ocupa un volumen de 16.3 cm

Answers

Answer:

[tex]d=785.27\ g/cm^3[/tex]

Explanation:

The question says that, "Calculate the absolute density of a fluid whose mass is 12.8 kg and which occupies a volume of 16.3 cm³".

Mass, m = 12.8 kg = 12800 g

Volume, V = 16.3 cm³

The density of an object is defined as the mass per unit volume i.e.

[tex]d=\dfrac{m}{V}\\\\d=\dfrac{12800}{16.3}\\\\d=785.27\ g/cm^3[/tex]

So, the absolute density of the fluid is equal to [tex]785.27\ g/cm^3[/tex].

What are some ways scientists can study the seafloor?

Answers

Answer: A device records the time it takes sound waves to travel from the surface to the ocean floor and back again. Sound waves travel through water at a known speed. Once scientists know the travel time of the wave, they can calculate the distance to the ocean floor.

Explanation:

can someone help me with this please what are the two correct answers??​

Answers

Answer:

The answers should be B and D

Explanation:

I hope this helps!

The answers are B and D

Astronomers build optical telescopes on tops of mountains because a. there is less air to dim the light. b. the seeing is better. c. CCDs work better with a beautiful mountain view. Astronomers build optical telescopes on tops of mountains because a. there is less air to dim the light. b. the seeing is better. c. CCDs work better with a beautiful mountain view. all choices, a, b and c choice c only choice a only choice b only choices a and b

Answers

Answer: choices a and b

Explanation:

Telescope can be defined as am optical instrument which is designed to observe the distant objects clear and nearer. It comprises of arrangement of lenses which allow the rays of light to be collected. The collected light is focused and the image so produced is magnified in the form of an image. The telescopes are prepared and manufactured on mountains top as this will help in preventing the distortion of light obtain from the star due to the fluctuation of air mass in the atmosphere. The atmospheric distortion affects the resolution, and affects the vision. The atmospheric pressure is low at the mountain tops so it will help in better observation of the sky.

A television that requires an average of 0.40 ampere of current is operated on a
120-volt service for 5.0 hours. How much energy is used?
a. 1.5 kWh
b. 0.15 kWh
C. 0.24 kWh
d. 0.67 kWh

Answers

The energy that has been used is 0.24 kWh


Explain how air exerts a pressure​

Answers

air exert a pressure take a glass and put the water on it and small piece a paper and glass take downword and observe it so tha air exert pressure

Which part of and organism is preserved in cast and mold fossils

Answers

Answer:

bones

Explanation:

Bones are is preserved in cast and mold fossils.

Normally the other 'parts of the body' like muscles, ligaments, blood and organs are soft.

Answer:

bones

Explanation:

I need help with this question it's science

Answers

What is the question?
I CANT SEE THE QUESTION

Someone help me extra points & brainlest

Answers

Answer:

75 m

Explanation:

(True/False) The normal force is always opposite in direction to weight.

Answers

Answer:

True

Explanation:

I WILL GIVE U BRAINLIST layers: Earth:: ___: the Sun
A: Milky Way galaxy
B: Zero gravity
C: Solid surface
D:Gas

Answers

Answer:

There are layers on the Earth.  There is gas on the Sun.

This means that the answer is...

D: Gas

CAN SOMEONE HELP ME WITH THIS

Answers

Answer:

D

Explanation:

Answer is D, you are welcome

What will be the change in velocity of a 850 kg car if a force of 50,000N is
applied for 0.5 s? (Round to one decimal)

Answers

Answer:

i'm not quite sure

Explanation:

just grive me a sec

Answer:

29.4 m/s

Explanation:

f = ma

a = [tex]\frac{f}{m}[/tex]

a = [tex]\frac{50000}{850}[/tex]

a = 58.8 [tex]m/s^{2}[/tex]

a = [tex]\frac{change in velocity}{t}[/tex]

change in velocity= at

change in velocity = 58.8 × 0.5

change in velocity = 29.4 m/s

Select the correct location on the image.
Which color in the visible spectrum has the highest frequency?

Answers

Answer:

Violet I took the test and it was right

Answer:

Violet the dude above was correct i took the test and got it right

Explanation:

A 4kg object has a momentum of 12 kg*m/s, what is the objects velocity?

Answers

Momentum = mass x velocity
12 = 4 x v | ÷ both sides by 4
12 ÷ 4 =v
v= 3 m/s

Marcus tries to move a refrigerator but can't. Has he done work?

A) Yes, because he used energy.
B) Yes, because he exerted a force.
C) No, because he did not move the object.
D) No, because he did not use energy.

Answers

Answer:

A

Explanation:

Even if the refrigerator did not move he still used energy to try to push it it's just his energy wasnt stronger than the refrigerator's energy

You pull straight up on the string of a yo-yo with a force 0.19 N, and while your hand is moving up a distance 0.12 m, the yo-yo moves down a distance 0.30 m. The mass of the yo-yo is 0.058 kg, and it was initially moving downward with speed 3.4 m/s. (b) What is the new speed of the yo-yo

Answers

Answer:

2.54 m/s

Explanation:

By conservation of energy,

work done by force due to hand = mechanical energy gained by yo-yo

So Fd = mgh + 1/2m(v₂² - v₁²)

where F = force on string = 0.19 N, d = distance moved by hand = 0.12 m, m = mass of yo-yo = 0.058 kg, g = acceleration due to gravity = 9.8 m/s², h = distance moved by yo-yo = 0.30 m, v₁ = initial speed of yo-yo = 3.4 m/s and v₂ = final speed of yo-yo = unknown

So,  Fd = mgh + 1/2m(v₂² - v₁²)

Fd/m = gh + 1/2(v₂² - v₁²)

Fd/m - gh = 1/2(v₂² - v₁²)

2(Fd/m - gh) = (v₂² - v₁²)

(v₂² - v₁²) = 2(Fd/m - gh)

v₂²  = v₁² + 2(Fd/m - gh)

taking square-root of both sides, we have

v₂  = √[v₁² + 2(Fd/m - gh)]

Substituting the values of the variables into the equation, we have

v₂  = √[(3.4 m/s)² + 2(0.19 N × 0.12 m/0.058 kg - 9.8 m/s² × 0.30 m)]

v₂  = √[(3.4 m/s)² + 2(0.19 N × 0.12 m/0.058 kg - 9.8 m/s² × 0.30 m)]

v₂  = √[(3.4 m/s)² + 2(0.0228 Nm/0.058 kg - 9.8 m/s² × 0.30 m)]

v₂  = √[(3.4 m/s)² + 2(0.393 Nm/kg - 2.94 m²/s²)]

v₂  = √[(3.4 m/s)² + 2(-2.547 m²/s²)]

v₂  = √[11.56 m²/s² - 5.094 m²/s²)]

v₂  = √[6.466 m²/s²]

v₂ = 2.54 m/s

A 90° elbow in a horizontal pipe is used to direct water flow upward at a rate of 40 kg/s. The diameter of the entire elbow is 10 cm. The elbow discharges water into the atmosphere, and thus the pressure at the exit is the local atmospheric pressure. The elevation difference between the centers of the exit and the inlet of the elbow is 50 cm. The weight of the elbow and the water in it is considered to be negligible. Determine (a) the gage pressure at the center of the inlet of the elbow and (b) the anchoring force needed to hold the elbow in place. Take the momentum-flux correction factor to be 1.03 at both the inlet and the outlet.

Answers

Answer:

(a) The gauge pressure is 4,885.3 Pa

(b) The anchoring force needed to hold the elbow in place is approximately 296.5 N

The direction of the anchoring force is approximately ≈ 134.8

Explanation:

The given parameters in the fluid dynamics question are;

Pipe elbow angle, θ = 90°

The mass flowrate of the water, [tex]\dot m[/tex] = 40 kg/s

The diameter of the elbow, D = 10 cm

The pressure at the point of discharge of the fluid = Atmospheric pressure

The elevation between the inlet and exit of the elbow, z = 50 cm

The weight of the elbow and the water = Negligible

(a) The velocity of the fluid, v₁ = v₂ = v is given as follows;

ρ·v·A = [tex]\dot m[/tex]

v = [tex]\dot m[/tex]/(ρ·A)

Where;

ρ = The density of the fluid (water) = 997 kg/m³

A = The cross-sectional area of the of the elbow = π·D²/4 = π×0.1²/4 ≈ 0.00785398163

A = 0.00785398163 m²

v = 40 kg/s/(0.00785398163 m² × 997 kg/m³) ≈ 5.1083 m/s

Bernoulli's equation for the flow of fluid is presented as follows;

[tex]\dfrac{P_1}{\rho \cdot g} +\dfrac{v_1}{2 \cdot g} + z_1 = \dfrac{P_2}{\rho \cdot g} +\dfrac{v_2}{2 \cdot g} + z_2[/tex]

v₁ = v₂ for the elbow of uniform cross section

P₁ - P₂ = ρ·g·(z₂ - z₁)

P₁ - P₂ = The gauge pressure = [tex]P_{gauge}[/tex]

z₂ - z₁ = z = 50 cm = 0.5 m

∴ [tex]P_{gauge}[/tex] = 997 kg/m³ × 9.8 m/s² × 0.5 m = 4,885.3 Pa

The gauge pressure, [tex]P_{gauge}[/tex] = 4,885.3 Pa

(b) The forces acting on the elbow are;

[tex]F_{Rx}[/tex] + [tex]P_{gauge}[/tex]·A = -β·[tex]\dot m[/tex]·v

[tex]F_{Ry}[/tex]  = β·[tex]\dot m[/tex]·v

∴ [tex]F_{Rx}[/tex]  = -β·[tex]\dot m[/tex]·v -  [tex]P_{gauge}[/tex]·A

[tex]F_{Rx}[/tex]  = -1.03 × 40 kg/s × 5.1083 m/s -  4,885.3 Pa × 0.00785398163 m² ≈ -208.83 N

[tex]F_{Ry}[/tex]  = 1.03 × 40 kg/s × 5.1083 m/s = 210.46196 N

The resultant force, [tex]F_R[/tex], is given as follows;

[tex]F_R[/tex] = √([tex]F_{Rx}[/tex]² + [tex]F_{Rz}[/tex]²)

∴ [tex]F_R[/tex] = √((-208.83)² + (210.46196)²) ≈ 296.486433934

Therefore;

The anchoring force needed to hold the elbow in place, [tex]F_R[/tex] ≈ 296.5 N

The direction of the anchoring force, θ

[tex]\theta = tan^{-1}\left( \dfrac{F_{Ry}}{F_{Rx}} \right)[/tex]

∴ θ = arctan(210.46196/(-208.83)) = -45..2230044°

θ = 180° + -45.2230044° = 134.7769956°

∴ The direction of the anchoring force is approximately, θ ≈ 134.8°

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