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Gnom [1K]
3 years ago
13

Which best describes the velocity of a person on a merry-go-round? * 1 point zero constant increasing continuously changing​

Chemistry
1 answer:
pav-90 [236]3 years ago
7 0

Answer:

continuously changing

Explanation:

The velocity of a person on a merry-go-round can best be described as continuously changing. This is because the initial velocity before the ride starts would be 0. Once the ride begins, it keeps going faster and faster until it reaches a certain speed. During this time the velocity is increasing. Then it remains constant for a certain period of time and when the ride is over the velocity and speed begins to decrease until ultimately reaching a speed of 0. Therefore, it is continuously changing throughout the duration of the ride.

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Write a hypothesis about the effect of dry conditions on earthworm behavior. Use the "if . . . then . . .
Vlada [557]

sorry about the late response...

<u>If an earthworm is exposed to dry conditions, then it will retreat to a moist place because its skin needs to stay moist for the earthworm to survive.</u>

4 0
3 years ago
Read 2 more answers
A person's heart beats 72 times a minute. At this rate, how many times does it beat in one hour?
sesenic [268]

At one minute, a persons's heart beats 72 times.

Therefore, in one hour or 60 minutes, a person's heart will beat (60×72) times, i.e., 4320 times

Answer is 4320 times

6 0
3 years ago
Naphthalene, C10H8, melts at 80.2°C. If the vapour pressure of the liquid is 1.3 kPa at 85.8°C and 5.3 kPa at 119.3°C, use th
sweet-ann [11.9K]

(a) One form of the Clausius-Clapeyron equation is

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:

  • P₁ = 1.3 kPa
  • P₂ = 5.3 kPa
  • T₁ = 85.8°C = 358.96 K
  • T₂ = 119.3°C = 392.46 K

Solving for ΔHv:

  • ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
  • ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
  • ΔHv = 49111.12 J/molK

(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:

  • ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
  • 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
  • 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
  • 1/T₂ = 2.049 * 10⁻³ K⁻¹
  • T₂ = 488.1 K = 214.94 °C

(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.

3 0
3 years ago
he heat of fusion of tetrahydrofuran is . Calculate the change in entropy when of tetrahydrofuran melts at . Be sure your answer
lana [24]

Answer:

\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}

Explanation:

Hello.

In this case, given the heat of fusion of THF to be 8.5 kJ/mol and freezing at -108.5 °C, for the required mass of 5.9 g, we can compute the entropy as:

\Delta S=\frac{n*\Delta H}{T}

Whereas n accounts for the moles which are computed below:

n=5.9g*\frac{1mol}{72g} =0.082mol

Thus, the entropy turns out:

\Delta S=\frac{0.0819mol*8.5 kJ/mol}{(-108.5+273.15)K}\\\\\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}

Best regards.

3 0
3 years ago
What is the ratio of effusion rates of krypton and neon at the same temperature and pressure?
egoroff_w [7]
The formula we use would be the graham's law. We do as follows:

<span>E_Kr / E_Ne = sqrt ( M_Ne / M_Kr) 
</span>
<span>= sqrt ( 20.1797 g/mol / 83.798 g/mol ) </span>
<span>= sqrt (0.24081) </span>
<span>= 0.4907
</span>
Hope this answers the question. Have a nice day.
7 0
3 years ago
Read 2 more answers
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