Heating and cooling can cause matter to change state. Which of the following occurs when heat is added to a system?
2 answers:
You might be interested in
Answer:
v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s
Explanation:
We can solve this problem using the kinematic relations, we have a three-dimensional movement, but we can work as three one-dimensional movements where the only parameter in common is time (a scalar).
X axis.
They indicate the initial position x = 8 m, its initial velocity v₀ = 5.5 m / s, the force Fx₁ = 220 N x₁ = 14 m, now the force changes to Fx₂ = 100 N up to the point xf = 17 m. The final speed is wondered.
As this movement is in three dimensions we must find the projection of the initial velocity in each axis, for this we can use trigonometry
the angle fi is with respect to the in z and the angle theta with respect to the x axis.
sin φ = z / r
Cos φ = r_p / r
z = r sin φ
r_p = r cos φ
the modulus of the vector r can be found with the Pythagorean theorem
r² = (x-x₀) ² + (y-y₀) ² + (z-z₀) ²
r² = (14-8) 2 + (-21 + 30) 2+ (-7 +4) 2
r = √126
r = 11.23 m
Let's find the angle with respect to the z axis (φfi)
φ = sin⁻¹ z / r
φ = sin⁻¹ ( )
φ = 15.5º
Let's find the projection of the position vector (r_p)
r_p = r cos φ
r_p = 11.23 cos 15.5
r_p = 10.82 m
This vector is in the xy plane, so we can use trigonometry to find the angle with respect to the x axis.
cos θ = x / r_p
θ = cos⁻¹ x / r_p
θ = cos⁻¹ ( )
θ = 56.3º
taking the angles we can decompose the initial velocity.
sin φ = v_z / v₀
cos φ = v_p / v₀
v_z = v₀ sin φ
v_z = 5.5 sin 15.5 = 1.47 m / z
v_p = vo cos φ
v_p = 5.5 cos 15.5 = 5.30 m / s
cos θ = vₓ / v_p
sin θ = v_y / v_p
vₓ = v_p cos θ
v_y = v_p sin θ
vₓ = 5.30 cos 56.3 = 2.94 m / s
v_y = 5.30 sin 56.3 = 4.41 m / s
we already have the components of the initial velocity
v₀ = (2.94 i ^ + 4.41 j ^ + 1.47 k ^) m / s
let's find the acceleration on this axis (ax1) using Newton's second law
Fₓx = m aₓ₁
aₓ₁ = Fₓ / m
aₓ₁ = 220/100
aₓ₁ = 2.20 m / s²
Let's look for the velocity at the end of this interval (vx1)
Let's be careful if the initial velocity and they relate it has the same sense it must be added, but if the velocity and acceleration have the opposite direction it must be subtracted.
vₓ₁² = v₀ₓ² + 2 aₓ₁ (x₁-x₀)
let's calculate
vₓ₁² = 2.94² + 2 2.20 (14-8)
vₓ₁ = √35.04
vₓ₁ = 5.92 m / s
to the second interval
they relate it to xf
aₓ₂ = Fₓ₂ / m
aₓ₂ = 100/100
aₓ₂ = 1 m / s²
final speed
v_{xf}² = vₓ₁² + 2 aₓ₂ (x_f- x₁)
v_{xf}² = 5.92² + 2 1 (17-14)
v_{xf} =√41.05
v_{xf} = 6.41 m / s
We carry out the same calculation for each of the other axes.
Axis y
acceleration (a_{y1})
a_{y1} = F_y / m
a_{y1} = 460/100
a_[y1} = 4.60 m / s²
the velocity at the end of the interval (v_{y1})
v_{y1}² = v_{oy}² + 2 a_{y1{ (y₁ -y₀)
v_{y1}2 = 4.41² + 2 4.60 (-21 + 30)
v_{y1} = √102.25
v_{y1} = 10.11 m / s
second interval
acceleration (a_{y2})
a_{y2} = F_{y2} / m
a_{y2} = 260/100
a_{y2} = 2.60 m / s2
final speed
v_{yf}² = v_{y1}² + 2 a_{y2} (y₂ -y₁)
v_{yf}² = 10.11² + 2 2.60 (-27 + 21)
v_{yf} = √ 71.01
v_{yf} = 8.43 m / s
here there is an inconsistency in the problem if the body is at y₁ = -27m and passes the position y_f = -21 m with the relationship it must be contrary to the velocity
z axis
first interval, relate (a_{z1})
a_{z1} = F_{z1} / m
a_{z1} = -200/100
a_{z1} = -2 m / s
the negative sign indicates that the acceleration is the negative direction of the z axis
the speed at the end of the interval
v_{z1}² = v_{zo)² + 2 a_{z1} (z₁-z₀)
v_{z1}² = 1.47² + 2 (-2) (-7 + 4)
v_{z1} = √14.16
v_{z1} = -3.76 m / s
second interval, acceleration (a_{z2})
a_{z2} = F_{z2} / m
a_{z2} = 210/100
a_{z2} = 2.10 m / s2
final speed
v_{fz}² = v_{z1}² - 2 a_{z2} | z_f-z₁)
v_{fz}² = 3.14² - 2 2.10 (-3 + 7)
v_{fz} = √6.94
v_{fz} = 2.63 m / s
speed is v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s
Answer:
Diagram 1, 3 and 4 can be explained with the phenomenon of refraction.
Refraction occurs when a ray of light crosses the interface between two mediums with different optical density: when this occurs, the ray of light is bent and its speed changes, according to Snell's law
where are the refractive index of the 1st and 2nd medium
are the angle that the incident ray and the refracted ray makes with the normal to the interface
In diagram, 1, the ray of light arrives perpendicularly to the interface, so it is refracted through the medium but it doesn't change its direction (only its speed).
In diagram 3, the ray of light is refracted twice: at the 1st interface and at the 2nd interface. In the 1st case, it goes from a medium with lower refractive index to a medium with higher refractive index (), this means that
, so the ray bends towards the normal. Vice-versa, in the 2nd case the ray goes from a medium with higher refractive index to a medium with lower refractive index (
), so it bends away from the normal (
).
In diagram 4, the ray of light is also refracted twice. The ray of light here acts exactly the same as in diagram 3, h
However, this time the 2nd interface is the opposite direction with respect to diagram 3, so in this case the ray of light at the 2nd interface bends in the opposite direction (still away from the normal).
Diagram 2 instead is an example of reflection, that occurs when a ray of light bounces off the interface between the two mediums, withouth entering the 2nd medium.
According to the law of reflection:
- The incoming ray, the reflected ray and the normal to the boundary are all in the same plane
- The angle of incidence is equal to the angle of reflection (both are measured relative to the normal to the boundary)
Therefore in this diagram, the ray of light hits the boundary at approx. 45 degrees from the normal, and then it is reflected back approximately at 45 degrees on the other side with respect to the normal.