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3 years ago

15

Which phase change is the result of particles moving more quickly? (2 points) liquid to gas liquid to solid gas to liquid gas to

solid

1 answer:

Alexxandr [17]3 years ago

4 0

Answer - Liquid to Gas

Liquid to gas is your answer. Remember, particles are hardly moving as a solid, and gradually increase in speed during phase changes. Solid>Liquid>Gas>Plasma

Solid being the slowest, Plasma the fastest.

<em>Hope this helps!</em>

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The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his

Alex17521 [72]

Answer:

22m/s

Explanation:

Mass, m=60 kg

Force constant, k=1300N/m

Restoring force, Fx=6500 N

Average friction force, f=50 N

Length of barrel, l=5m

y=2.5 m

Initial velocity, u=0

$F_x=kx$

Substitute the values

$6500=1300x$

$x=\frac{6500}{1300}=5$ m

Work done due to friction force

$W_f=fscos\theta$

We have $\theta=180^{\circ}$

Substitute the values

$W_f=50\times 5cos180^{\circ}$

$W_f=-250J$

Initial kinetic energy, Ki=0

Initial gravitational energy, $U_{grav,1}=0$ \

Initial elastic potential energy

$U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)$

$U_{el,1}=16250J$

Final elastic energy, $U_{el,2}=0$

Final kinetic energy, $K_f=\frac{1}{2}(60)v^2=30v^2$

Final gravitational energy, $U_{grav,2}=mgh=60\times 9.8\times 2.5$

Final gravitational energy, $U_{grav,2}=1470J$

Using work-energy theorem

$K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}$

Substitute the values

$0+0+16250-250=30v^2+1470+0$

$16000-1470=30v^2$

$14530=30v^2$

$v^2=\frac{14530}{30}$

$v=\sqrt{\frac{14530}{30}}$

$v=22m/s$

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A ball of mass m= 450.0 g traveling at a speed of 8.00 m/s impacts a vertical wall at an angle of θi =45.00 below the horizontal

Mkey [24]

Answer:

F = 20.4 i ^

Explanation:

This exercise can be solved using the ratio of momentum and amount of movement.

I = F t = Dp

Since force and amount of movement are vector quantities, each axis must be worked separately.

X axis

Let's look for speed

cos 45 = vₓ / v

vₓ = v cos 45

vₓ = 8 cos 45

vₓ = 5,657 m / s

We write the moment

Before the crash p₀ = m vₓ

After the shock $p_{f}$ = -m vₓ

The variation of the moment Δp = mvₓ - (-mvₓ) = 2 m vₓ

The impulse on the x axis Fₓ t = Δp

Fₓ = 2 m vₓ / t

Fx = 2 0.450 5.657 / 0.250

Fx = 20.4 N

We perform the same calculation on the y axis

sin 45 = vy / v

vy = v sin 45

vy = 8 sin 45

vy = 5,657 m / s

We calculate the initial momentum po = m $v_{y}$

Final moment $p_{f}$ = m $v_{y}$

Variations moment Δp = m $v_{y}$ - m $v_{y}$ = 0

Force in the Y-axis $F_{y}$ = 0

Therefore the total force is

F = fx i ^ + Fyj ^

F = Fx i ^

F = 20.4 i ^

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