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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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If a single circular loop of wire carries a current of 61 A and produces a magnetic field at its center with a magnitude of 1.70
Schach [20]

Answer:

The  radius is  R =  0.22 5 \  m

Explanation:

From the question we are told that

    The current is  I  =  61 \ A

     The  magnetic field is  B  =  1.70 *10^{-4} \  T

Generally the magnetic field produced by a current carrying conductor  is mathematically represented as

        B  =  \frac{\mu_o  *  I}{2 *  R }

=>     R  =  \frac{\mu_o  *  I  }{ 2 *  B }

Here  \mu_o is the permeability of free space with value  \mu_o  =  4\pi * 10^{-7} N/A^2

=>    R  =  \frac{  4\pi * 10^{-7}  *   61  }{ 2 *   1.70 *10^{-4} }

=>  R =  0.22 5 \  m

8 0
3 years ago
One mole of a gas is placed in a closed system with a 20 L vessel initially at T = 300 K. The vessel is then isothermally expand
Elden [556K]

Answer:

Given that

P = RT/V + a/V²

We know that

H= U + PV

For T= Constant  (ΔU=0)

ΔH= ΔU +Δ( PV)

ΔH= Δ( PV)

P = RT/V + a/V²

P V= RT + a/V

dH/dV = d(RT + a/V)/dV

dH/dV = - a/V²

So the expression of dH/dV

\dfrac{dH}{dV}=\dfrac{-a}{V^2}

b)

In isothermal process

\Delta H=nRT\ln{\dfrac{V_2}{V_1}}      (ΔU=0)

Now by putting the all values

\Delta H=nRT\ln{\dfrac{V_2}{V_1}}

\Delta H=1\times 0.08206\times 300\ln{\dfrac{40}{20}}

ΔH = 17.06 L.atm

8 0
3 years ago
A body of mass 500g accelerates at a rate of 5 metres per second square. Calculate the force applied to the body ​
kenny6666 [7]

Answer:

2.5 N

Explanation:

F = m•a

F = 0.5•5

F = 2.5 N

5 0
3 years ago
If you were on the open ocean on a large ship, what steps would you do to determine the height of a wave?
Daniel [21]
The first step to take is to check the anemometer to determine the intensity of the wind blowing in the open ocean. The measurement is in terms of knots or nautical miles. Next, we use a Beaufort scale to check what range the knots falls in. Using the scale, we can see what is the height of the wave. The Beaufort scale is used in open oceans only
3 0
3 years ago
Se apunta un rifle horizontalmente con mira a un blanco pequeño que está a 200m en el suelo. La velocidad inicial de la bala es
vfiekz [6]

Answer:

Lo importante a tener en cuenta sobre esta pregunta es que la velocidad horizontal de la bala no hace ninguna diferencia en cuanto al tiempo que tarda en caer al suelo.

Debido a que el arma no ha aplicado ninguna fuerza vertical a la bala, la única fuerza que afecta la bala es la gravedad. Esto significa que la bala tarda tanto en caer al suelo como lo haría si se cayera, a pesar de que ahora viaja una gran distancia horizontal en la duración.

Para encontrar el tiempo de viaje antes de tocar el suelo, tenemos 3 valores:

-El desplazamiento desde el suelo que la bala debe viajar, s = 1.5m

-La aceleración que experimenta la bala. Como la gravedad está acelerando la bala hacia abajo, a = g = ~ 9.81m / s ^ 2

-La velocidad inicial de la bala verticalmente. Como la bala es estacionaria verticalmente (solo viaja horizontalmente al inicio), u = 0m

Examinamos nuestras ecuaciones de movimiento, comúnmente conocidas como ecuaciones SUVAT. Es posible que necesite aprender estos para su examen, pero algunas tablas de examen los proporcionan.

Debido a que tenemos s, u y a, y estamos buscando el tiempo t, la ecuación relevante es

s = ut + 0.5 (en ^ 2)

Completando nuestros valores tenemos:

1.5 = 0t + 0.5 (9.81 x t ^ 2)

1.5 = 4.905 x t ^ 2

Divide 1.5 entre 4.905 para encontrar t ^ 2

t ^ 2 = 0.3058 ...

Simplemente encontramos la raíz cuadrada de t ^ 2 para encontrar t, el tiempo que tarda la bala en llegar al suelo:

t = 0.553s (3 cifras significativas)

Para encontrar la distancia horizontal, d, que la bala ha viajado antes de tocar el suelo, podemos usar la ecuación que vincula el desplazamiento s con cierta velocidad v durante un tiempo t:

s = vt

La velocidad horizontal de la bala, v = 430

El tiempo antes de que la bala toque el suelo, t = 0.553

Entonces d = vt = 430 * 0.553 = 238m (3 cifras significativas)

3 0
4 years ago
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