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Marysya12 [62]
2 years ago
11

Writing an Argument About Galetown’s Severe Storms Note: If you do not have access to Amplify Science at home, your teacher will

provide you with an alternate way to complete this homework. In your final report to the citizens of Galetown, you will discuss the three claims and explain how all three factors can contribute to severe storms. Then, you will predict if the storms will always be severe. What caused Galetown to have more severe rainstorms this summer than in previous years?
Claim 1: The lake that was built near Galetown caused it to have more severe rainstorms.

Claim 2: Warmer weather caused Galetown to have more severe rainstorms.

Claim 3: Stronger winds caused Galetown to have more severe rainstorms.

Be sure to use some of the vocabulary words you have learned in both of your writing assignments: Word Bank air parcel, cloud,condensation, energy, evaporation, temperature, transfer, troposphere, water vapor, weather, wind   

Recall from the previous activity which claims you think are best supported by evidence. Use these claims to explain what is happening in Galetown. You can use the data table and the words listed in the word bank above to help you with your report that answers the question: What caused Galetown to have more severe rainstorms this summer than in previous years? Press NEXT to move on to the second part of the written report. Write a 3-5 paragraph argumented essay ​

Physics
1 answer:
jarptica [38.1K]2 years ago
4 0

The claim that can be deduced is that B. Warmer weather caused Galetown to have more severe rainstorms.

<h3>What is a claim?</h3>

It should be noted that a claim simply means the stance of an author in a literary work.

In this case, the claim is that warmer weather caused Galetown to have more severe rainstorms.

When the temperature of an air parcel begins higher, this will make it rise higher in the troposphere before the temperature of the air parcel and surrounding air are equal.

Learn more about claim on:

brainly.com/question/2748145

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During takeoff, an airplane climbs with a speed of 195 m/s at an angle of 15° above the horizontal. The speed and angle constitu
matrenka [14]

Answer:

The horizontal component of the velocity is 188 m/s

The vertical component of the velocity is 50 m/s.

Explanation:

Hi there!

Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:

We can find vx using the following trigonometric rule of a right triangle:

cos α = adjacent / hypotenuse

cos 15° = vx / 195 m/s

195 m/s · cos 15° = vx

vx = 188 m/s

The horizontal component of the velocity is 188 m/s

To calculate the y-component we will use the following trigonometric rule:

sin α = opposite / hypotenuse

sin 15° = vy / 195 m/s

195 m/s · sin 15° = vy

vy = 50 m/s

The vertical component of the velocity is 50 m/s.

4 0
3 years ago
La ecuación de la posición de una esferita está dada por: r(t)=(2.Cos(πt) i-3.Sen(πt) j) (m) ¿Cuál es la velocidad de la esferit
katovenus [111]

Answer:

v = (-4.44 i^ + 6.66 j^ )  m/s, a_average =( 0 i^ -2π j^) m/s²

Explanation:

The expression left corresponds to an oscillatory movement (MAS), the speed is defined by

         v = dr / dt

the function of position

         r = 2 cos πt  i^  + 3 sin πt  j^

let us note that it is a movement in two dimensions

let's perform the derivative

          v = -2π sin πt  i^  + 3π cos πt  j^

we evaluate this expression for t = 0.25 s, remember that the angle is in radians

          v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^

           v = (-4.44 i^ + 6.66 j^ )  m/s

To calculate the mean acceleration we use the expression

           a = (v_{f} - v_{o}) / Δt

 

indicates that the time is the first 3 s

       

we look for the initial velocity t = 0 s

           v₀ = 0 i ^ + 3π j ^

         

we look for the fine velocity, t = 3 s

          v_f = - 2π sin (π 3) + 3π cos (π 3) j ^

          v_f = 0 i ^ - 3π j ^

we calculate the average acceleration

            Δt = (3 -0) = 3 s

           a_average = (0-0) / 3 i ^ + (-3π - 3π) / 3

           a_average = (0 i ^ -2π j ^ ) m/s²

6 0
3 years ago
The ground state energy of an oscillating electron is 1.23 eV. How much energy must be added to the electron to move it to the t
Vikki [24]

Answer:

  • The energy that must be added to the electron to move it to the third excited state is  -1.153 eV
  • The energy that must be added to the electron to move it to the fourth excited state is  -1.181 eV

Explanation:

Given;

Energy of electron in ground state (n = 1 ) = 1.23 eV

E₁ = 1.23 eV

Eₙ = E₁ /n²

where;

E₁ is the energy of the electron in ground state

n is the energy level,

For third excited state, n = 4

E₄ = E₁ /4²

E₄ = (1.23 eV) / 16

E₄ = 0.077 eV

Change in energy level, = E₄ - E₁ = 0.077 eV - 1.23 eV = -1.153 eV

The energy that must be added to the electron to move it to the third excited state is  -1.153 eV

For fourth excited state, n = 5

E₅ = E₁ /5²

E₄ = (1.23 eV) / 25

E₄ = 0.049 eV

Change in energy level, = E₅ - E₁ = 0.049 eV - 1.23 eV = -1.181 eV

The energy that must be added to the electron to move it to the fourth excited state is  -1.181 eV

5 0
3 years ago
Four point charges have equal magnitudes. Three are positive, and one is negative, as the drawing shows. They are fixed in place
Semmy [17]

Answer:

Ans= 9

See attached picture for clearer solution.

Explanation:

The net electrostatic force acting on charge A = 2/ 2 + 2 /(2) 2 − 2 /(3) 2 = 2 / 2 (1 + 1/4 – 1/9 ) = 41/36 2/2 .  

The net electrostatic force acting on charge B = 2/2 + 2/(2)2 − 2/2 = 1/4 2/d2 .  

The net electrostatic force acting on charge C = 2/2 + 2/(2)2 + 2/2 = 2/2 (1 + 1 4 + 1) = 9/4 2/2 .  

The net electrostatic force acting on charge D = 2/2+ 2 /(2)2 + 2/(3)2 = 2 /2 (1 + 1/4 + 1/9 ) = 49/36 2/ 2 .  

The ratio of the largest to the smallest net force = 9/4*2/2  /  1/4 2/2 . = 9

5 0
3 years ago
3. Maverick and Goose are flying a training mission in their F-14. They are
Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of y_0=1500m, and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

y=y_0 - \frac{gt^2}{2}    [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

x = v.t     [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

t=\sqrt{\frac{2y_0}{g}}

Since we have y_0=1500m

t=\sqrt{\frac{2(1500)}{9.8}}

t=17.5 sec

Replacing in [2]

x = 688\ m/sec \ (17.5sec)

x = 12040\ m

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

6 0
3 years ago
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