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3 years ago

Which graph shows the relationship between temperature, X, and kinetic energy, Y?

Physics

2 answers:

strojnjashka [21]3 years ago

5 0

Answer: see the graph attached (straight line, passing through the origin and positive slope).

Justification:

1) Kinetic energy and temperature are in direct proportion. That means:

i) Being kinetic energy y and temperature x: y α x

ii) That implies: y = kx,where k is the constant of proportionality.

iii) The graph is a line that passes through the origin and has positive slope k (k = y / x).

2) The proportional relationship between kinetic energy (KE) and temperature (T) is shown by the Boltzman law, which states:

Average KE = [3 / 2] KT, where K is Boltzman's constant, whose graph is of the form shown in the figure attached.

prisoha [69]3 years ago

5 0

Answer:

c (third graph)

Explanation:

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A centrifuge accelerates uniformly from rest to 15000 rpm in 330 s . Through how many revolutions did it turn in this time?

eduard

Answer:

The number of revolutions turned by the centrifuge is 8250 revolutions.

Explanation:

Given;

number of revolution per minutes, ω = 15000 rpm

time of motion, t = 330 s = 5.5 minutes

The number of revolutions turned by the centrifuge is given by;

$N = \frac{1500 \ Rev}{minutes} *5.5 \ minutes\\\\N = 8250 \ revolutions$

Therefore, the number of revolutions turned by the centrifuge is 8250 revolutions.

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3 years ago

Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second

vampirchik [111]

To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.

From the law of Malus intensity can be defined as

$I' = \frac{I_0}{2} cos^2 \theta$

Where

$\theta =$ Angle From vertical of the axis of the polarizing filter

$I_0 =$ Intensity of the unpolarized light

The expression for the intensity of the light after passing through the first filter is given by

$I = \frac{I_0}{2}$

Replacing we have that

$I = \frac{370}{2}$

$I = 185W/m^2$

Re-arrange the equation,

$I'= \frac{I_0}{2}cos^2\theta$

Re-arrange to find \theta

$cos^2\theta = \frac{2I'}{I_0}$

$cos^2\theta = \frac{2*138}{370}$

$\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})$

$\theta = 0.5282rad$

$\theta = 30.27\°$

The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°

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3 years ago

Why are multiple images seen when two plane mirrors are placed at an angle

Whitepunk [10]

Answer: Can I get a picture???

8 0

3 years ago

The two equal strong kids are having a tug a war. What do you expect to happen to the ball in this situation

borishaifa [10]

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As such,The tension is equal to the applied force.The tension is trasmitted to the opposite side and of the rope delivering the applied force.

Hope this helps.. :)

3 0

3 years ago

if a person can jump maximum along distance of 3m ,on the earth how far could be jump on the moon where acceleration due to grav

allochka39001 [22]

Answer:

The person can jump 48 m on the Moon

Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, $R_{max}$ = 3 m

The acceleration due to gravity on the Moon = 1 ÷ 16 of that on Earth

The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, $R_{max}$ , of a projectile motion is found by using the following formula

$R_{max} = \dfrac{u^2}{g}$

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore, we have;

$R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }$

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, $g_{Moon}$ = 1/16 × g

∴ $g_{Moon}$ = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

$R_{max \ Moon} = \dfrac{u^2}{g_{Moon}} = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m$

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is $R_{max \ Moon}$ ≈ 48 m

8 0

3 years ago