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3 years ago

Can the tangent line to a velocity vs. time graph ever be vertical? Explain.

Physics

1 answer:

34kurt3 years ago

7 0

Answer:

No. Because it would correspond to zero Instantaneous acceleration.

Explanation:

hope this helps

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A balloon is filled to a volume of 7.00*10^2 mL at a temperature of 20.0°C. The balloon is then cooled at constant pressure to a

wolverine [178]

The final volume of the gas is 238.9 mL

Explanation:

We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):

$\frac{V}{T}=const.$

Which can be also re-written as

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where

$V_1, V_2$ are the initial and final volumes of the gas

$T_1, T_2$ are the initial and final temperature of the gas

For the gas in the balloon in this problem, we have:

$V_1 = 7.00\cdot 10^2 mL = 700 mL$ is the initial volume

$T_1=20.0^{\circ}C+273=293 K$ is the initial absolute temperature

$V_2$ is the final volume

$T_2 = 1.00\cdot 10^2 K = 100 K$ is the final temperature

Solving for $V_2$ ,

$V_2 = \frac{V_1 T_2}{T_1}=\frac{(700)(100)}{293}=238.9 mL$

Learn more about ideal gases:

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6 0

3 years ago

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

$\Delta P.E=mg(h_{2}-h_{1})$

$\Delta P.E=10000\times9.8\times(10000-0)$

$\Delta P.E=10000\times9.8\times10000$

$\Delta P.E=980000000\ J$

$\Delta P.E=980\ MJ$

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)$

$\Delta K.E=148298642\ J$

$\Delta K.E=148.3\ MJ$

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

7 0

3 years ago