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4 years ago

Can I put an “I” for an argumentative essay?

Physics

1 answer:

masha68 [24]4 years ago

4 0

Answer:

There is no place for first person in formal academic argumentative essay. However, you can use first person only if it has been asked to provide your personal opinion or reflection about the topic.

Explanation:

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Hi can someone pls answer the question number 2 pls i srsly need it. thank u :)

Anvisha [2.4K]

The distance between the image and its leans is 19.04 cm...

Refer to the attachment for detailed solution!!~

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2 years ago

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On top of Mount Everest, the temperature is -19 ∘C in July. Being a physicist, you determine by how many degrees Celsius one nee

kondaur [170]

Answer:

254 °C

Explanation:

The average kinetic energy of gas molecules K = 3RT/2N where R = gas constant = 8.314 J/mol-K, N = avogadro's constant = 6.022 × 10²³ atoms/mol

T = temperature in Kelvin.

Let K be its average kinetic energy at t = -19°C = 273 + (-19) = 273 - 19 = 254 K = T. K = 3RT/2N = 3 × 8.314 J/mol-K × 254 K/(2 × 6.022 × 10²³ atoms/mol) = 5.26 × 10⁻²¹ J

When its average kinetic energy doubles, it becomes K₁ = 2K = 2 × 5.26 × 10⁻²¹ = 10.52 × 10⁻²¹ J at temperature T₂. So,

K₁ = 3RT₁/2N

T₁ = 2NK₁/3R

T₁ = 2 × 6.022 × 10²³ atoms/mol × 10.52 × 10⁻²¹ J/3 × 8.314 J/mol-K = 508 K

The temperature difference is thus ΔT = T₁ - T = 508 K - 254 K = 254 K.

Since temperature change in kelvin scale equals temperature change in Celsius scale ΔT = 254 °C

So, we need to change the temperature of the air by 254 °C to double its average kinetic energy.

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3 years ago

7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr

Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

$d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m$

$d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m$

From the continuity equations in pipes we have to

$A_1V_1 = A_2 V_2$

Where,

$A_{1,2}$ = Cross sectional Area at each section

$V_{1,2}$ = Flow Velocity at each section

Then replacing we have,

$(\pi r_1^2) v_1 = (\pi r_2^2) v_2$

$(1.25*10^{-2})^2 v_1 = ( 0.06*10^{-2})^2 v_2$

$v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1$

From Bernoulli equation we have that the change in the pressure is

$\Delta P = \frac{1}{2} \rho (v_2^2-v_1^2)$

$7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)$

$7300= 8919.01 v_1^2$

$v_1 = 0.9m/s$

Therefore the speed of flow in the first tube is 0.9m/s

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4 years ago

Which of the following processes could NOT be used to separate dissolved particles from the liquid in a solution? *

DIA [1.3K]

Answer:

Vaporation

Explanation:

In the vaporization or boiling, the passage of particles from the liquid state to the gaseous state occurs completely

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4 years ago

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State in which atoms or molecules are very close together and are regularly arranged

umka21 [38]

Solid, because the atoms maintain their form and do not take the shape of their container. The particles share tight, close bonds due to this set shape

7 0

3 years ago

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