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german
2 years ago
9

The absolute temperature of a gas is t. in order to double the rms speed of its molecules, what should be the new absloute tempe

rature?
Physics
1 answer:
jenyasd209 [6]2 years ago
6 0

The new absloute temperature should be 4t.

<h3>Temperature </h3>

The hotness of matter or radiation is expressed by the physical quantity known as temperature.

There are three different types of temperature scales: those, like the SI scale, that are defined in terms of the average translational kinetic energy per freely moving microscopic particle, like an atom, molecule, or electron in a body; those that solely depend on strictly macroscopic properties and thermodynamic principles, like Kelvin's original definition; and those that are not defined by theoretical principles but rather by useful empirical properties of particula.

Using a thermometer, one can gauge temperature. It is calibrated using different temperature scales, each of which historically defined itself using a different set of reference points and thermometric materials.

Learn more about temperature here:

brainly.com/question/15267055

#SPJ4

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If Scobie could drive a Jetson's flying car at a constant speed of 160.0 km/hr across oceans and space, approximately how long w
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To calculate the time that takes Scobie to drive around Earth's equator we need to find the distance, which is given by the equation of a circumference:

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Then, the distance is:

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6 0
3 years ago
a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long
Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

a=\frac{Vf-Vi}{t}

-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

-9.8  =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2

To solve for "t" in this quadratic equation, we can factor it out as shown:

0= t(20-\frac{9.8}{2} t)

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s

7 0
3 years ago
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