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balu736 [363]
2 years ago
13

In three to four sentences, explain how climate change could impact the formation of hurricanes. (4 points)

Physics
1 answer:
daser333 [38]2 years ago
5 0

The climate change can directly impact the formation of hurricanes because increasing temperature increases intensity of hurricanes.

<h3>How climate change could impact the formation of hurricanes?</h3>

Evaporation increases as temperatures rise which also causes the transfer of heat from the oceans to the atmosphere. As the storms travel across oceans whose water is warm, they absorb more water vapor and heat which increases rainfall as well as intensity of hurricanes.

So we can conclude that the climate change impacted the formation of hurricanes due to increasing temperature that increases intensity of hurricanes.

Learn more about climate here: brainly.com/question/17922964

You might be interested in
A sample of monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to
leonid [27]

Answer:

(a) 0.203 moles

(b) 900 K

(c) 900 K

(d) 15 L

(e) A → B, W = 0, Q = Eint = 1,518.91596 J

B → C, W = Q ≈ 1668.69974 J Eint = 0 J

C → A, Q = -2,531.5266 J, W = -1,013.25 J, Eint = -1,518.91596 J

(g) ∑Q = 656.089 J, ∑W =  655.449 J, ∑Eint = 0 J

Explanation:

At point A

The volume of the gas, V₁ = 5.00 L

The pressure of the gas, P₁ = 1 atm

The temperature of the gas, T₁ = 300 K

At point B

The volume of the gas, V₂ = V₁ = 5.00 L

The pressure of the gas, P₂ = 3.00 atm

The temperature of the gas, T₂ = Not given

At point C

The volume of the gas, V₃ = Not given

The pressure of the gas, P₃ = 1 atm

The temperature of the gas, T₂ = T₃ = 300 K

(a) The ideal gas equation is given as follows;

P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

n = The number of moles present

R = The universal gas constant = 0.08205 L·atm·mol⁻¹·K⁻¹

n = PV/(R·T)

∴ The number of moles, n = 1 × 5/(0.08205 × 300) ≈ 0.203 moles

The number of moles in the sample, n ≈ 0.203 moles

(b) The process from points A to B is a constant volume process, therefore, we have, by Gay-Lussac's law;

P₁/T₁ = P₂/T₂

∴ T₂ = P₂·T₁/P₁

From which we get;

T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K

The temperature at point B, T₂ = 900 K

(c) The process from points B to C is a constant temperature process, therefore, T₃ = T₂ = 900 K

(d) For a constant temperature process, according to Boyle's law, we have;

P₂·V₂ = P₃·V₃

V₃ = P₂·V₂/P₃

∴ V₃ = 3.00 atm. × 5.00 L/(1.00 atm.) = 15 L

The volume at point C, V₃ = 15 L

(e) The process A → B, which is a constant volume process, can be carried out in a vessel with a fixed volume

The process B → C, which is a constant temperature process, can be carried out in an insulated adjustable vessel

The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston

(f) For A → B, W = 0,

Q = Eint = n·cv·(T₂ - T₁)

Cv for monoatomic gas = 3/2·R

∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J

Q = Eint = 1,518.91596 J

For B → C, we have a constant temperature process

Q = n·R·T₂·㏑(V₃/V₂)

∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J

Eint = 0

Q = W ≈ 1668.69974 J

For C → A, we have a constant pressure process

Q = n·Cp·(T₁ - T₃)

∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J

Q = -2,531.5266 J

W = P·(V₂ - V₁)

∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J

W = -1,013.25 J

Eint = n·Cv·(T₁ - T₃)

Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J

Eint = -1,518.91596 J

(g) ∑Q = 1,518.91596 J + 1668.69974 J - 2,531.5266 J = 656.089 J

∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J

∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J

5 0
3 years ago
The higher the temperature of an object the
Allushta [10]

 higher temp = higher energy = higher frequency = shorter wavelength

4 0
3 years ago
A steady electric current flows through a wire. If 9.0 C of charge passes a particular spot in the wire in a time period of 2.0
defon

1) Current: 4.5 A

2) Time taken: 4.7 s

Explanation:

1)

The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:

I=\frac{q}{t}

where

I is the current

q is the amount of charge passing a given point in a time t

For the wire in this problem, we have

q = 9.0 C is the amount of charge

t = 2.0 s is the time interval

Solving for I, we find the current:

I=\frac{9.0}{2.0}=4.5 A

2)

To solve this problem, we can use again the same formula

I=\frac{q}{t}

where

I is the current

q is the amount of charge passing a given point in a time t

In this problem, we have:

I = 3.0 A (current)

q = 14.0 C (charge)

Therefore, the time taken for the charge to move past a particular spot in the wire is

t=\frac{q}{I}=\frac{14.0}{3.0}=4.7 s

Learn more about electric current:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

8 0
3 years ago
A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.
Airida [17]

Answer:

a) v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} ), b) \Delta K = 9.559\times 10^{-5}\,J

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v

The final velocity is:

v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )

b) The change in the kinetic energy of the 13.5 g coin is:

\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]

\Delta K = 9.559\times 10^{-5}\,J

4 0
3 years ago
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2. do not dry wash

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