Diagram-A satisfies. High amplitude (bright) and long wavelengths are present on the left (red). & The right side has a short wavelength and low amplitude (dim) (violet).
Light comes from a source as waves. Each wave has an electric and a magnetic component. Light is hence sometimes referred to as electromagneticradiation.
A large portion of the light in the universetravels with wavelengths that are too short or too long for the human eye to detect, yet our brains interpret light waves by giving distinct colours to the various wavelengths.
The infrared, microwave, and radio spectrum bands have the longest wavelengths. The ultraviolet, x-ray, and gammaradiation have the shortest wavelengths in the electromagnetic spectrum.
Diagram A is therefore satisfactory. On the left, there are long wavelengths with high amplitude (bright) (red). & The right side is dark and has a short wavelength (violet).
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Answer:
Δ KE = - 8.75 x 10⁻⁴ J
Explanation:
given,
mass of applesauce = 7 g = 0.007 Kg
initial velocity, u = 0.5 m/s
final velocity, v = 0 m/s
Decrease in kinetic energy = ?
initial kinetic energy
KE₁ = 8.75 x 10⁻⁴ J
final kinetic energy
KE₂ =0 J
Decrease in kinetic energy
Δ KE = KE₂ - KE₁
Δ KE = 0 - 8.75 x 10⁻⁴
Δ KE = - 8.75 x 10⁻⁴ J
decrease in kinetic energy of the applesauce is equal to 8.75 x 10⁻⁴ J
The decrease in kinetic energy of the applesauce, when it hits the wall and stops, is the initial kinetic energy of it. Using the formula of kinetic energy, the decrease is calculated to be 0.000875 Joules.
This question relates to the concept of kinetic energy in physics. Kinetic energy is calculated by the formula 0.5 * mass (kg) * velocity (m/s)^2. So the initial kinetic energy of the applesauce right after being thrown was 0.5 * 0.007 kg * (0.5 m/s)^2 = 0.000875 Joules.
When the applesauce hits the wall and stops, its velocity drops to 0. Thus, its kinetic energy also goes to 0 (because kinetic energy is proportional to the square of velocity).
Therefore, the decrease in kinetic energy is the same as the initial kinetic energy of the applesauce, which is 0.000875 Joules.
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This function might represent the lateral displacement of a string, a local electric field, the position of the surface of a body of water, or any of a number of other physical manifestations of waves.
1. Find ye(x) and yt(t). Keep in mind that yt(t) should be a trigonometric function of unit amplitude.
2. At the position x=0, what is the displacement of the string (assuming that the standing wave ys(x,t) is present)?
3. At certain times, the string will be perfectly straight. Find the first time t1>0 when this is true.
4. Which one of the following statements about the wave described in the problem introduction is correct?
A. The wave is traveling in the +x direction.
B. The wave is traveling in the −x direction.
C. The wave is oscillating but not traveling.
D. The wave is traveling but not oscillating.
Which of the expressions given is a mathematical expression for a wave of the same amplitude that is traveling in the opposite direction? At time t=0this new wave should have the same displacement as y1(x,t), the wave described in the problem introduction.
A. Acos(kx−ωt)
B. Acos(kx+ωt)
C. Asin(kx−ωt)
D. Asin(kx+ωt)
The definition of standing wave and trigonometry allows to find the results for the questions about the waves are:
1. For the standing wave its parts are: spatial and
temporal part
2. The string moves with an oscillating motion y = A’ cos wt.
3. Thefirst displacement is zero for
4. the correct result is:
A. The wave is traveling in the +x direction.
5. The correct result is:
D. Asin(kx+ωt)
Traveling waves are periodic movements of the media that transport energy, but not matter, the expression to describe it is:
y₁ = A sin (kx -wt)
Where A is the amplitude of the wave k the wave vector, w the angular velocity and x the position and t the time.
1. Ask us to find the spatial and temporal part of the standing wave.
To form the standing wave, two waves must be added, the reflected wave is:
y₂ = A sin (kx + wt)
The sum of a waves
y = y₁ + y₂
y = A (sin kx-wt + sin kx + wt)
We develop the sine function and add.
Sin (a ± b) = sin a cos b ± sin b cos a
The result is:
y = 2A sin kx cos wt
They ask that the function be unitary therefore
The amplitude of each string
A_ {chord} = A_ {standing wave} / 2
The spatial part is
= A 'sin kx
The temporary part is:
= A ’cos wt
2. At position x = 0, what is the displacement of the string?
y = A ’cos wt
The string moves in an oscillating motion.
3. At what point the string is straight.
When the string is straight its displacement is zero x = 0, the position remains.
y = A ’cos wt
For the amplitude of the chord to be zero, the cosine function must be zero.
wt = (2n + 1)
the first zero occurs for n = 0
wt =
t =
4) The traveling wave described in the statement is traveling in the positive direction of the x axis, therefore the correct statement is:
A. The wave is traveling in the +x direction.
5) The wave traveling in the opposite direction is
y₂ = A sin (kx + wt)
The correct answer is:
D. Asin(kx+ωt)
In conclusion using the definition of standing wave and trigonometry we can find the results for the questions about the waves are:
1. For the standing wave its parts are: spatial and
temporal part
2. The string moves with an oscillating motion y = A’ cos wt.
3. Thefirst displacement is zero for
4. the correct result is:
A. The wave is traveling in the +x direction.
5. The correct result is:
D. Asin(kx+ωt)
Learn more about standing waves here: brainly.com/question/1121886
To solve the problem it is necessary to apply the concepts related to thermal expansion of solids. Thermodynamically the expansion is given by
Where,
Original Length of the bar
= Change in temperature
= Coefficient of thermal expansion
On the other hand our values are given as,
Replacing we have,
The width of the expansion of the cracks between the slabs is 0.5832cm
The width of the expansion cracks between the slabs to prevent buckling should be 0.5832cm.
According to this question, the following information are given:
The values are given as follows:
∆L = Loα (T2 - T1)
∆L = 18 × 12 × 10-⁶ (27)
∆L = 3.24 × 10-⁴ × 18
∆L = 5.832 × 10-³m
Therefore, the width of the expansion of the cracks between the slabs is 0.5832cm.
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Answer:
5N
Explanation:
(25 N - 20 N = 5 N)
Answer:
speeds above 343 m/s
Explanation:
I have taken the test got 100%