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galben [10]
2 years ago
10

_________ refers to the process whereby at low pressures a solid changes directly into the vapor phase without passing through t

he liquid phase.
Physics
1 answer:
Ira Lisetskai [31]2 years ago
6 0
Sublimation <span>refers to the process whereby at low pressures a solid changes directly into the vapor phase without passing through the liquid phase~

</span>
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A boat takes off from a dock at 2.5 m/s and speeds up at 4.2 m/s squared for six seconds how far has the most traveled
GaryK [48]

The boat's position x relative to its starting point x_0=0 is determined by

x=x_0+v_0t+\dfrac12at^2

where v_0 is its initial velocity, a is its acceleration, and t is time. After t=6\,\mathrm s, the boat has traveled

x=\left(2.5\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(4.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2

\implies x=91\,\mathrm m

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What is capital of Thaivan?
lisabon 2012 [21]

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2 years ago
How do determine which is Y2, Y1, X2, and X1 on a graph. <br><br> And how do find rise over run.
fredd [130]
One point will be X1,Y1 and the other will be X2,Y2. It does not matter which is which except that X1 and Y1 have to be the same point and X2 and Y2 have to be the same point. For example, let's say you were given (2,3) and (6,8). No matter which point is X1,Y1 and the other is X2,Y2, the slope will still be 5/4. 

The rise is the change in y from one point to the other. The run would be the change in x from one point to the other.
7 0
3 years ago
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
3 years ago
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