Answer:
a. Microwaves—3 and infrared waves—1
Explanation:
Microwaves and infrared waves are both part of the electromagnetic spectrum, but they have different frequency and wavelength.
In particular:
- Microwaves are long-wavelength electromagnetic waves, with wavelength between 1 mm and 1 m. Their wavelength is longer than visible light
- Infrared waves are also long-wavelength electromagnetic waves, but their wavelength is shorter than microwaves: between 700 nm and 1 mm. Their wavelength is also longer than visible light.
The two types of waves are also used for different purposes. In particular:
- Infrared waves are emitted by any hot object, and their intensity depends on the temperature of the object. Therefore, they are used in astronomy to show the heat released by astronomical objects (option 1)
- Microwaves are used to study the Cosmic Microwave Background (CMB). This is electromagnetic radiation that permeates the whole universe, and its wavelength depends inversely on the local temperature. Therefore, areas with longer wavelength have lower temperature, and viceversa. Therefore, microwaves are used to measure temperature differences in space (option 3).
This effect is explained by increased chain entanglements at higher molecular weights. Increasing the degree of crystallinity of a semicrystalline polymer leads to an enhancement of the tensile strength. Deformation by drawing increases the tensile strength of a semicrystalline polymer.
Answer:
Explanation:
Puck A's initial speed is 
and move in a direction 
after the collision.
#
since there's no external force on the system(
#The collision equation can be written as;
The kinetic energies before and after the collision are expressed as:
Let +x along A's initial direction and +y along A's final direction makes the angle 
#Substitute in 
:
Hence, the the speed of puck A after the collision is 12.891 m/s
#. The velocity of A after the collision is;
Substitute 
into 
:
This is the velocity of puck B after the collision, it's speed is:
The velocity of puck B after the collision is 6.8544 m/s
c. The direction of puck B after the collision is:
Hence, the direction of B's velocity after the collision is 62°
Answer:
t = 3.82 s
Ax = 147 N (←)
Ay = 294 N (+↑)
Explanation:
Given
m = 30.0 Kg
ωinitial = 125 rad/s
ωfinal = 0 rad/s
μC = 0.5
R = 0.3 m
t = ?
Ax = ?
Ay = ?
For the disk, we can apply
∑ τ = I*α
where I = m*R²/2
then
⇒ R*Ffriction = (m*R²/2)*α
⇒ R*(-μC*N) = R*(-μC*m*g) = (m*R²/2)*α
⇒ α = -2*μC*g / R
⇒ α = -2*(0.5)*(9.8) / 0.3 = -32.666 rad/s²
we can use the equation to get t:
α = Δω / t ⇒ t = Δω / α = (0 - 125) / (-32.666)
⇒ t = 3.82 s
The horizontal and vertical components of force which the member AB exerts on the pin at A during this time are
∑ Fx = 0 (+→)
Ax - Ffriction = 0
⇒ Ax = Ffriction = μC*m*g = (0.5)*(30)*(9.8) = 147 N
⇒ Ax = 147 N (←)
∑ Fy = 0 (+↑)
⇒ Ay - m*g = 0
⇒ Ay = m*g
⇒ Ay = 30*9.8 = 294 N
⇒ Ay = 294 N (+↑)
Answer:
Wind deposits sand into a small mound. So the answer is Deposition
