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3 years ago

9

5. The equation shows the formation of nitrogen and hydrogen. What effect does increasing the pressure have on this equilibrium?

Hint: which reaction produces fewer gas molecules?

1 answer:

bezimeni [28]3 years ago

6 0

more hydrogen and nitrogen form

Hope this helps :)

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You drop a rock down a well and hear a splash 3 s later. As Charlie Brown would say, the well is 3 seconds deep. How fast is the

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Since it was dropped, it should be the speed of gravity which is 9.8 meters/second

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2 years ago

:) What is practical machine? what is the reda<br>tion between MA and VR in a practical<br>machine?

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Answer: For ideal machine efficiency = 1. Hence M.A = V. R. The V. R of an ideal machine and the practical machine is a constant or is the same for both

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A hiker climbs a mountain. Starting at the base of the mountain, he first moved up 520m at a 32.0 degree angle. What is the fina

balu736 [363]

Answer:

$\displaystyle \vec{d}=$

Explanation:

<u>Displacement Vector</u>

Suppose an object is located at a position

$\displaystyle P_1(x_1,y_1)$

and then moves at another position at

$\displaystyle P_2(x_2,y_2)$

The displacement vector is directed from the first to the second position and can be found as

$\displaystyle \vec{d}=$

If the position is given as magnitude-angle data ( z , α), we can compute its rectangular components as

$\displaystyle x=z\ cos\alpha$

$\displaystyle y=z\ sin\alpha$

The question describes the situation where the initial point is the base of the mountain, where both components are zero

$\displaystyle P_1(0,0)$

The final point is given as a 520 m distance and a 32-degree angle, so

$\displaystyle x_2=520\ cos32^o= 440.99\ m$

$\displaystyle y_2=520\ sin32^o=275.6\ m$

The displacement is

$\displaystyle \vec{d}=$

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3 years ago

The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c

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To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

$X_f = X_i +\frac{1}{2}(V_i+V_f)t$

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

$X_f = \frac{1}{2} (V_i+V_f)t$

Clearing for time,

$t = \frac{2X_f}{(V_i+V_f)}$

$t = \frac{2*1.2}{24+0}$

$t = 0.1s$

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

$a= \frac{\Delta V}{t}$

Then

$F = m\frac{\Delta V}{t}$

$F = m\frac{V_f-V_i}{t}$

$F = m\frac{-V_i}{t}$

$F = \frac{(1600kg)(-24m/s)}{(0.1s)}$

$F = -384000N$

Negative symbol is because the force is opposite of the direction of moton.

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$V_f^2=V_i^2-2ax$

$0 = (24m/s)^2-2*a(1.2m)$

$a = \frac{(24m/s)^2}{1.2m}$

$a=480m/s^2$

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$a = 53.3g$

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