Refraction.
L<span>ight behavior in which light goes through a medium and bends, going in a different direction.</span>
Reflection.
Light behavior in which light waves bounce off a medium and goes in a different direction.
Concave.
A surface shaped in such a way that the center curves inward like this:
Convex.
A surface shaped in such a way that the center curves outward like this:
I just took the test I hope this helps. :)
Answer:
Speed of the plane will be equal to 2.969 m/sec
Explanation:
Note - It is not given but we have to find the speed of the plane.
We have given radius r =
Angular acceleration
We have to fond the speed of the plane
We know that angular acceleration is equal to
So
v = 2.969 m/sec
So speed of the plane will be equal to 2.969 m/sec
Answer:
Coording metal is a physical change.
Answer:
N = 136.77 N
Explanation:
This is an exercise in Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis. In the attachment we can see the applied forces.
Let's use trigonometry to decompose the force F1
cos θ = F₁ₓ / F₁
sin θ = F_{1y} / F₁
F₁ₓ = F₁ cos θ
F_{1y} = F₁ sin θ
now let's apply Newton's second law to each axis
X axis
F₁ₓ - F4 = m a
Y axis
N + F3 + F_{1y} -F₂ -W = 0
the acceleration can be calculated with kinematics
v = v₀ + a t
since the object starts from rest, the initial velocity is zero v₀ = 0
a = v / t
a = 20/3
a = 6.667 m / s²
we substitute in the equation
F₁ₓ = F₄ + m a
F₁ₓ = 42 + 15 6,667
F₁ₓ = 142 N
F₁ cos θ = 142
cos θ = 142/206 = 0.6893
θ = cos⁻¹ 0.6893
θ = 46.42º
now let's work the y axis
N = W + F₂ - F₃ - F_{1y}
N = 15 9.8 + 144 -5 - 206 sin 46.42
N = 286 - 149.23
N = 136.77 N
Answer:
0.0389 cm
Explanation:
The current density in a conductive wire is given by
where
I is the current
A is the cross-sectional area of the wire
In this problem, we know that:
- The fuse melts when the current density reaches a value of
- The maximum limit of the current in the wire must be
I = 0.62 A
Therefore, we can find the cross-sectional area that the wire should have:
We know that the cross-sectional area can be written as
where d is the diameter of the wire.
Re-arranging the equation, we find the diameter of the wire: