Answer: (1) The correct answer is A.
(2) The correct answer is D.
Explanation:
(1)
Reflection is the sending back of light from the surface without absorbing it. In the reflection phenomenon, the wave does not continue moving forward.
Diffraction is the bending of the light around the obstacle. In the diffraction phenomenon, the wave travels forward after striking around the obstacle.
Therefore, the correct answer is A.
(2)
Amplitude is the maximum displacement in the medium from the rest position.
The amount of energy is related to the amplitude. Amplitude is related to the amount of energy carried by the wave. Low energy wave is characterized by a low amplitude. High energy wave is characterized by a high amplitude.
Therefore, the correct option is D.
Answer:
225 N
Explanation:
"Below the horizontal" means he's pushing down at an angle.
Draw a free body diagram of the box. There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.
Sum of forces in the y direction:
∑F = ma
N − mg − F sin θ = 0
N = F sin θ + mg
Plug in values:
N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)
N = 225 N
Answer:
Density is affected by volume and mass.
Explanation:
Density is defined as the quantity of mass per unit of volume, or expressed mathematically, d = m/v.
So the area under a velocity time graph is distance or displacement, if you have done calculus yet you will understand that if you take the integral of a velocity function then you end up with displacement. Thats for later understanding however.
So this appears to be a right triangle so we can find the area of a triangle as:
0.5bh = A
Since our area is 10 meters lets alter our formula a bit to fit the situation:
Our base here is time and our height is velocity so:
0.5tv = Δx
So we can read off the graph that our velocity at the end, or our final velocity appears to be near 2.0 m/s
So we have v, and Δx so lets isolate for time by dividing by v and 0.5
t = Δx / 0.5v
Now lets plug all that in:
t = 10 / 0.5(2)
t = 10 seconds
Hope this helped!
Answer:
62.8 μC
Explanation:
Here is the complete question
The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?
Solution
The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ
So, Q = ∫∫∫ρdV
Q = ∫∫∫ρr²sinθdθdrdΦ
Q = ∫∫∫(0.2r²)r²sinθdθdrdΦ
Q = ∫∫∫0.2r⁴sinθdθdrdΦ
We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π
So, Q = ∫∫∫0.2r⁴sinθdθdrdΦ
Q = ∫∫∫0.2r⁴[∫sinθdθ]drdΦ
Q = ∫∫0.2r⁴[-cosθ]drdΦ
Q = ∫∫0.2r⁴-[cosπ - cos0]drdΦ
Q = ∫∫∫0.2r⁴-[-1 - 1]drdΦ
Q = ∫∫0.2r⁴-[- 2]drdΦ
Q = ∫∫0.2r⁴(2)drdΦ
Q = ∫∫0.4r⁴drdΦ
Q = ∫0.4r⁴dr∫dΦ
Q = ∫0.4r⁴dr[Φ]
Q = ∫0.4r⁴dr[2π - 0]
Q = ∫0.4r⁴dr[2π]
Q = ∫0.8πr⁴dr
Q = 0.8π∫r⁴dr
Q = 0.8π[r⁵/5]
Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]
Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]
Q = 0.8π[0.025 m⁵ - 0 m⁵]
Q = 0.8π[0.025 m⁵]
Q = (0.02π mC/m⁵) m⁵
Q = 0.0628 mC
Q = 0.0628 × 10⁻³ C
Q = 62.8 × 10⁻³ × 10⁻³ C
Q = 62.8 × 10⁻⁶ C
Q = 62.8 μC
