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asambeis
8 days ago
7

A city planner needs to verify that Main Street and Oak Street are parallel. If ___________ are supplementary, the streets must

be parallel. A) ∠1 and ∠2 B) ∠1 and ∠3 C) ∠2 and ∠3 D) ∠2 and ∠4
Physics
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Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta
ValentinkaMS [3465]

Answer:

The amount of heat that enters the gas throughout this two-step process totals 120 cal.

Explanation:

Given that,

Moles present = 3

Heat capacity at volume held constant = 4.9 cal/mol.K

Heat capacity at pressure held constant = 6.9 cal/mol.K

Starting temperature = 300 K

Ending temperature = 320 K

We are tasked with determining the heat absorbed by the gas at constant pressure

Employing the heat formula

\Delta H_{1}=nC_{p}\times\Delta T

Substituting the values into the equation

\Delta H_{1}=3\times6.9\times(320-300)

\Delta H_{1}=414\ cal

Next, we calculate the heat absorbed by the gas at constant volume

Using the corresponding heat formula

\Delta H_{1}=nC_{v}\times\Delta T

Insert the values into the formula

\Delta H_{1}=3\times4.9\times(300-320)

\Delta H_{1}=-294\ cal

Now, it's necessary to evaluate the total heat flow into the gas during both steps

Using the total heat formula

\Delta H_{T}=\Delta H_{1}+\Delta H_{2}

\Delta H_{T}=414-294

\Delta H_{T}=120\ cal

Thus, the heat that transfers into the gas throughout this two-step process amounts to 120 cal.

7 0
3 months ago
During a snowball fight two balls with masses of 0.4 and 0.6 kg, respectively, are thrown in such a manner that they meet head-o
Yuliya22 [3333]

Answer:

The snowball's speed after the impact is 3 m/s

Explanation:

Given the following:

mass of each ball

m₁ = 0.4 Kg

m₂ = 0.6 Kg

initial speed of both balls = v₁ = 15 m/s

Speed of 1 Kg mass post-collision =?

Applying conservation of momentum

m₁ v₁ - m₂ v₁ = (m₁+m₂) V

A negative velocity indicates that the second ball moves in the opposite direction.

0.4 x 15 - 0.6 x 15 = (1) V

Therefore,

V = - 3 m/s

Consequently,

The snowball's speed following the collision is 3 m/s

8 0
3 months ago
The G string on a guitar is 69 cm long and has a fundamental frequency of 196 Hz. A guitarist can play different notes by pushin
Keith_Richards [3271]
Δd = 23 cm. When the eta string of the guitar has nodes at both ends, the resulting waves create a standing wave, which can be expressed with the following formulas: Fundamental: L = ½ λ, 1st harmonic: L = 2 ( λ / 2), 2nd harmonic: L = 3 ( λ / 2), Harmonic n: L = n λ / 2, where n is an integer. The rope's speed can be calculated using the formula v = λ f. This speed remains constant based on the tension and linear density of the rope. Now, let's determine the speed with the provided data: v = 0.69 × 196, yielding v = 135.24 m/s. Next, we will find the wavelengths for the two frequencies: λ₁ = v / f₁, which gives λ₁ = 135.24 / 233.08, equaling λ₁ = 0.58022 m; λ₂ = v / f₂ results in λ₂ = 135.24 / 246.94, consequently λ₂ = 0.54766 m. We'll substitute into the resonance equation Lₙ = n λ/2. At the third fret, m = 3, therefore L₃ = 3 × 0.58022 / 2, resulting in L₃ = 0.87033 m. For the fourth fret, m = 4, which gives L₄ = 4 × 0.54766 / 2, equating to L₄ = 1.09532 m. The distance between the two frets is Δd = L₄ – L₃, so Δd = 1.09532 - 0.87033, leading to Δd = 0.22499 m or 22.5 cm, rounded to 23 cm.
6 0
2 months ago
A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or
Softa [3030]

Answer:

 Baseball mass:  m_b=3.992*10^{14}kg  

Explanation:

 Circumference of a baseball is calculated using 2πr = 23 cm

 Thus, the radius comes out to be 3.66 cm, which equals 3.66*10^{-2} m

 The mass density of the baseball matches that of a neutron or proton.

 Proton mass = 10^{-27} kg  

 Proton diameter = 10^{-15} m

 Proton radius =  5*10^{-16} m

 Volume of the baseball is \frac{4}{3} \pi r^3

 Now by substituting all values into the mass per unit volume equation for the baseball, we get:  

         \frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}

         \frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}  

         m_b=3.992*10^{14}kg

       Therefore, the baseball mass amounts to m_b=3.992*10^{14}kg              

5 0
3 months ago
Read 2 more answers
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