Answer:
When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.
Explanation:
As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.
Blueshifts happens when the source of the wave and the observer are moving closer towards one another.
Assume that the wave is of frequency [tex]f\; {\rm Hz}[/tex] at the source. In other words, the source of the wave sends out a peak after every [tex](1/f)\; {\text{seconds}}[/tex].
Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.
The source of this wave sends out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. It would appear to the observer that consecutive peaks arrive every [tex](1/f)\; {\text{seconds}}\![/tex]. That would correspond to a frequency of [tex]f\; {\rm Hz}[/tex].
On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of [tex]v \; {\rm m \cdot s^{-1}}[/tex].
Again, the source of this wave would send out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. However, by the time the source sends out the second peak, the source would have been [tex]v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}[/tex] closer to the observer then when the source sent out the first peak.
When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than [tex](1/f)\; {\text{seconds}}[/tex]. The observed frequency of this wave would be larger than the original [tex]f\; {\rm Hz}[/tex].
A car hurtles off a cliff and crashes on the canyon floor below. Identify the system in which the net momentum is zero during the crash.
Solution :
It is given that a car ran off from a cliff and it crashes on canyon floor. Now the system of a car as well as the earth together have a [tex]\text{ net momentum of zero}[/tex] when the car crashes on the canyon floor, thus reducing the momentum of the car to zero. The earth also stops its upward motion and it also reduces the momentum to zero.
A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity increases to 57 rad/s. Assume that the angular acceleration was constant during this time interval. How many revolutions does the wheel turn through during this time interval
Answer:
The number of revolutions is 44.6.
Explanation:
We can find the revolutions of the wheel with the following equation:
[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]
Where:
[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s
t: is the time = 8 s
α: is the angular acceleration
We can find the angular acceleration with the initial and final angular velocities:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Where:
[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s
[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]
Hence, the number of revolutions is:
[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]
Therefore, the number of revolutions is 44.6.
I hope it helps you!
A nerve impulse travels along a myelinated neuron at 90.1 m/s.
What is this speed in mi/h?
Answer:
201.5537 mph
Explanation:
Given the following data;
Speed = 90.1 m/s
Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.
Mathematically, speed is given by the formula;
Speed = distance/time
To convert this value into miles per hour;
Conversion;
1 meter = 0.000621 mile
90.1 meters = 90.1 * 0.000621 = 0.05595 miles
1 metre per second = 2.237 miles per hour
90.1 meters per seconds = 90.1 * 2.237 = 201.5537 miles per hour
90.1 m/s = 201.5537 mph
A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frictional force of 2 N, find (a) its acceleration, and (b) its initial velocity.
A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.60 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)
Express your answer using two significant figures.
Answer:
0.5
Explanation:
because the block is attached to the pulley of the string
two identical eggs are dropped from the same height. The first eggs lands on a dish and breaks, while the second lands on a pillow and does not break. Which quantities are the same in both situations
Answer:
The height is the same
Explanation:
Because they were at the same height but they fell at different velocities
A cylindrical disk of wood weighing 45.0 N and having a diameter of 30.0 cm floats on a cylinder of oil of density 0.850 g>cm3 (Fig. E12.19). The cylinder of oil is 75.0 cm deep and has a diameter the same as that of the wood. (a) What is the gauge pressure at the top of the oil column
Answer:
665.25 Pa
Explanation:
Given data :
Weight of the disk, w = 45 N
Diameter, d = 30 cm
= 0.30 m
Therefore, radius of the disk,
[tex]$r=\frac{d}{2}$[/tex]
[tex]$r=\frac{0.30}{2}$[/tex]
= 0.15 m
Now, area of the cylindrical disk,
[tex]$A=\pi r^2$[/tex]
[tex]$A=3.14 \times (0.15)^2$[/tex]
[tex]$=0.07065 \ m^2$[/tex]
∴ The gauge pressure at the top of the oil column is :
[tex]$p=\frac{w}{A}$[/tex]
[tex]$p=\frac{47}{0.07065}$[/tex]
= 665.25 Pa
Therefore, the gauge pressure is 665.25 Pa.
The definition of pressure allows to find the result for the pressure at the top of the oil cylinder is:
The pressure is: P = 636.6 Pa
The pressure is defined by the relationship between perpendicular force and area.
[tex]P = \frac{F}{A}[/tex]
where P is pressure, F is force, and A is area.
They indicate that the wooden cylinder weighs W = 45.0 N and has a diameter of d = 30 cm = 0.30 m.
The area is:
A = π r² = [tex]\pi \frac{d^2}{4}[/tex]
In the attachment we see a diagram of the forces, where the weight of the cylinder and the thrust are equal.
B-W = 0
B = W
The force applied to the liquid is the weights of the cylinder. Let's replace.
[tex]P= \frac{W}{A} \\P = W \frac{4}{\pi d^2 }[/tex]
Let's calculate.
[tex]P = \frac{45 \ 4 }{\pi \ 0.30^2 }[/tex] P = 45 4 / pi 0.30²
P = 636.6 Pa
In conclusion using the definition of pressure we can find the result for the pressure at the top of the oil cylinder is:
The pressure is: P = 636.6 Pa.
Learn more about pressure here: brainly.com/question/17467912
After de Broglie proposed the wave nature of matter, Davisson and Germer demonstrated the wavelike behavior of electrons by observing an interference pattern from electrons scattering off what
Answer:
Scattering is an interaction that can happen when a given particle or wave, like an electron, impacts a target or material. Then the electron changes it's original path and leaves some energy in the process. (This is a really simplified explanation of scattering, this is a really complex phenomenon, but let's not dive into that path)
Particularly, Davisson and Germer used a beam of electrons against a target of nickel, and these scattered electrons were detected by a detector. All of that in a vacuum chamber.
Then the correct answer is a nickel target.
"After de Broglie proposed the wave nature of matter, Davisson and Germer demonstrated the wavelike behavior of electrons by observing an interference pattern from electrons scattering off a nickel target"
Traveling waves propagate with a fixed speed usually denoted as v (but sometimes c). The waves are called __________ if their waveform repeats every time interval T.
a. transverse
b. longitudinal
c. periodic
d. sinusoidal
Answer:
periodic
Explanation:
how can scientific method solve real world problems examples
3. You have a variable-voltage power supply and a capacitor in the form of two metal disks of radius 0.6 m, held a distance of 1 mm apart. What is the largest voltage you can apply to the capacitor without the air becoming highly conductive
Answer:
The breakdown of air occurs at a maximum voltage of 3kV/mm.
Explanation:
The breakdown of air occurs at a maximum voltage of 3kV/mm.
At this level of voltage the air between the plates become highly ionised and breakdown occurs. Since, the distance held between the plates is 1mm , it can withstand a maximum voltage of 3 kV.
After this voltage the air will become conductive in nature and will form ions in the air between the plates and ultimately breakdown will take place with a flash.
The density of blood is 1055 kg/m3 . If the blood at the very top of your head exerts a minimum gauge pressure of 45 mm Hg (6000 Pa), estimate the gauge pressure at your heart in pascals.
Answer:
P = 10135.6 Pa
Explanation:
For this exercise we use that the pressure varies with the height
P = P₀ + ρ g h
where h is the height from the head to the heart, which is approximately
h = 40 cm = 0.40m and P₀ is the head pressure P₀ = 6000 Pa
P = 6000 + 1055 9.8 0.40
P = 6000 + 4135.6
P = 10135.6 Pa
Question 7 of 10
A railroad freight car with a mass of 32,000 kg is moving at 2.0 m/s when it
runs into an at-rest freight car with a mass of 28,000 kg. The cars lock
together. What is their final velocity?
A.1.1 m/s
B. 2.2 m/s
C. 60,000 kg•m/s
D. 0.5 m/s
Answer:
a
Explanation:
you take 32,000kg ÷2.0m
1. A 20.0 N force directed 20.0° above the horizontal is applied to a 6.00 kg crate that is traveling on a horizontal
surface. What is the magnitude of the normal force exerted by the surface on the crate?
N = 52.0 N
Explanation:
Given: [tex]F_a= 20.0\:\text{N}=\:\text{applied\:force}[/tex]
[tex]m=6.00\:\text{kg}[/tex]
[tex]N = \text{normal force}[/tex]
The net force [tex]F_{net}[/tex] is given by
[tex]F_{net} = N + F_a\sin 20 - mg=0[/tex]
Solving for N, we get
[tex]N = mg - F_a\sin 20[/tex]
[tex]\:\:\:\:\:\:= (6.00\:\text{kg})(9.8\:\text{m/s}^2) - (20.0\:\text{N}\sin 20)[/tex]
[tex]\:\:\:\:\:\:= 52.0\:\text{N}[/tex]
uppose that 3 J of work is needed to stretch a spring from its natural length of 32 cm to a length of 49 cm. (a) How much work (in J) is needed to stretch the spring from 37 cm to 45 cm
Answer:
0.113 J
Explanation:
Applying,
w = ke²/2................. Equation 1
Where w = workdone in stretching the spring, k = spring constant, e = extension
make k the subject of the equation
k = 2w/e²................ Equation 2
From the question,
Given: w = 3 J, e = 49-32 = 17 cm = 0.17 m
Substitute these values into equation 2
k = (2×3)/0.17²
k = 6/0.17
k = 35.29 N/m
(a) if the spring from 37 cm to 45 cm,
Then,
w = ke²/2
Given: e = 45-37 = 8 cm = 0.08
w = 35.29(0.08²)/2
w = 0.113 J
It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the spring an additional 0.10 m, does it take 130 J, more than 130 J or less than 130 J? Verify your answer with a calculation.
Explanation:
Given that,
Work done to stretch the spring, W = 130 J
Distance, x = 0.1 m
(a) We know that work done in stretching the spring is as follows :
[tex]W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m[/tex]
(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m
So,
[tex]W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J[/tex]
So, the new work is more than 130 J.
The slope of a d vs t graph represents velocity. Describe 3 ways you know this to be true.
Answer:
Look at explanation
Explanation:
I only know 1 way, there is another way you can rephrase this using derivatives but that's pretty much the same thing.
The slope is calculated by Δy/Δx so the slope of distance vs time graph is Δd/Δt which is the velocity
2.
Select the correct answer.
Erica is working in the lab. She wants to remove the fine dust particles suspended in a sample of oil. Which method is she most likely to use?
Answer:
Reverse Osmosis
Explanation:
Reverse osmosis is a type of filtration that involves passing a solvent through a semipermeable membrane in the opposite direction that natural osmosis does. Separation is always enforced through the use of pressure in this process. Ions, fine dust particles, molecules, and larger particles are typically removed from solvents using this method. The technique is particularly popular in the treatment and purification of water.
Answer:
filtration is used to separate
You walk into a room and you see 4 chickens on a bed 2 cows on the floor and 2 cats in a chair. How many legs are on the ground? (I know this answer just a riddle to see who knows it) (:
Answer:
18
Explanation:
I'm pretty sure I got it right
A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are vertical. String A is attached at a distance d
Answer:
a) T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] , b) T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]
c) x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex], d) m₂ = m₁ ( [tex]\frac{ L}{2d} -1[/tex])
Explanation:
After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act
a) The tension of string A is requested
The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive
∑ τ = 0
T_A d - W₂ x -W₁ L/2 = 0
T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]
b) the tension in string B
we write the expression of the translational equilibrium
∑ F = 0
T_A - W₂ - W₁ - T_B = 0
T_B = T_A -W₂ - W₁
T_ B = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] - g m₂ - g m₁
T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]
c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system
T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0
at the point that begins to rotate T_B = 0
g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0
m₂ (d-x) = m₁ (0.5 L- d)
m₂ x = m₂ d - m₁ (0.5 L- d)
x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex]
d) The mass of the block for which it is always in equilibrium
this is the mass for which x = 0
0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}
[tex]\frac{m_1}{m_2} \ (0.5L -d) = d[/tex]
[tex]\frac{m_1}{m_2} = \frac{ d}{0.5L-d}[/tex]
m₂ = m₁ [tex]\frac{0.5 L -d}{d}[/tex]
m₂ = m₁ ( [tex]\frac{ L}{2d} -1[/tex])
A 6.0-cm-diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?
Answer:
a n c
Explanation:
Which of the following statements is correct about the magnitude of the static friction force between an object and a surface?
a. Static friction depends on the mass of the object.
b. Static friction depends on the shape of the object.
c. Static friction depends on what the object is made of but not what the surface is made of.
d. None of the above is correct.
Answer:
Static friction depends on the mass of the object.
Explanation:
Friction is the force between two surfaces in contact. The force of friction between two surfaces in contact depends on;
1) nature of the object and the surface(how rough or smooth the surfaces are)
2)surface area of the object and the surface
3) mass of the object
Since;
F=μmg
Where;
μ= coefficient of static friction
m= mass of the object
g= acceleration due to gravity
Hence, as the mass of the object increases, the magnitude of static friction force between an object and a surface increases and vice versa.
A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward in 1.71 s. He breaks the tackle and runs straight forward another 24.0 m in 5.20 s. Calculate his average velocity (in m/s) for the entire motion. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
Answer:
Average Velocity = 3.63 m/s
Explanation:
First, we will calculate the total displacement of the quarterback, taking forward direction as positive:
Total Displacement = 15 m - 3 m + 24 m = 36 m
Now, we will calculate the total time taken for this displacement:
Total Time = 3 s + 1.71 s + 5.2 s = 9.91 s
Therefore, the average velocity will be:
[tex]Average\ Velocity = \frac{Total\ Displacement}{Total\ Time}\\\\Average\ Velocity = \frac{36\ m}{9.91\ s}[/tex]
Average Velocity = 3.63 m/s
No esporte coletivo, um dos principais fatores desenvolvidos é o desenvolvimento social. Qual desses não faz parte das virtudes ensinadas no esporte?
Companheirismo
Humildade
Ser justo (Fair Play)
Vencer independente do que precise ser feito
Answer:
fair palybtgshsisuehdh
A wave moves in a rope with a certain wavelength. A second wave is made to move in the same rope with twice the wavelength of the first wave. The frequency of the second wave is _______________ the frequency of the first wave.
Answer:
The frequency of the second wave is half of the frequency of first one.
Explanation:
The wavelength of the second wave is double is the first wave.
As we know that the frequency is inversely proportional to the wavelength of the velocity is same.
velocity = frequency x wavelength
So, the ratio of frequency of second wave to the first wave is
[tex]\frac{f_2}{f_1} =\frac{\lambda _1}{\lambda _2}\\\\\frac{f_2}{f_1} =\frac{\lambda _1}{2\lambda _1}\\\\\frac{f_2}{f_1} =\frac{1}{2}\\\\[/tex]
The frequency of the second wave is half of the frequency of first one.
If a bale of hay behind the target exerts a constant friction force, how much farther will your arrow burry itself into the hay than the arrow from the younger shooter
Answer:
The arrow will bury itself farther by 3S₁
Explanation:
lets assume; the Arrow shot by me has a speed twice the speed of the arrow fired by the younger shooter
Given that ; acceleration is constant , Frictional force is constant
A₂ = A₁
Vf²₂ - Vi²₂ / 2s₂ = Vf₁² - Vi₁² / 2s₁ ---- ( 1 )
final velocities = 0
Initial velocities : Vi₂ = 2(Vi₁ )
Back to equation 1
0 - (2Vi₁ )² / 2s₂ = 0 - Vi₁² / 2s₁
hence :
s₂ = 4s₁
hence the Arrow shot by me will burry itself farther by :
s₂ - s₁ = 3s₁
Note : S1 = distance travelled by the arrow shot by the younger shooter
During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race
Answer:
The maximum speed of Bolt for the 100 m race is 14.66 m/s
Explanation:
Given;
initial distance covered by Bolt, d = 200 m
time of this motion, t = 19.3 s
The second distance covered by Bolt, = 100 m
Assuming Bolt maintained the same acceleration for both races.
His acceleration can be determined from the 200 m race.
d = ut + ¹/₂at²
where;
u is his initial velocity = 0
d = ¹/₂at²
[tex]at^2 = 2d\\\\a = \frac{2d}{t^2} \\\\a = \frac{2\times 200}{19.3^2} \\\\a = 1.074 \ m/s^2[/tex]
Let the final or maximum velocity for the 100 m race = v
v² = u² + 2ad₂
v² = 2 x 1.074 x 100
v² = 214.8
v = √214.8
v = 14.66 m/s
The maximum speed of Bolt for the 100 m race is 14.66 m/s
Two pistons are connected to a fluid-filled reservoir. The first piston has an area of 3.002 cm2, and the second has an area of 315 cm2. If the first cylinder is pressed inward with a force of 50.0 N, what is the force that the fluid in the reservoir exerts on the second cylinder?
Answer:
The force on the second piston is 5246.5 N .
Explanation:
Area of first piston, a = 3.002 cm^2
Area of second piston, A = 315 cm^2
Force on first piston, f = 50 N
let the force of the second piston is F.
According to the Pascal's law
[tex]\frac{f}{a} = \frac{F}{A}\\\\\frac{50}{3.002}=\frac{F}{315}\\\\F = 5246.5 N[/tex]
a girl is moving with a uniform velocity of 1.5 m/s then mathematically find her acceleration
Answer:
0
Explanation:
a = dv/dt
if v is constant than the slope of the v graph will be 0, so dv/dt is 0
a= 0
ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on its circular path is 6.0 m/s. What is its kinetic energy at an instant when the string makes an angle of 50 degree with the vertical
Answer:
K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
Em₀ = K = ½ mv²
final point. When it is at tea = 50º
Em_f = K + U
Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
h = L - L cos 50
Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
Em₀ = Em_f
½ mv² = ½ m v_b² + mg L (1 - cos50)
K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
K_b = 36 +42.0
K_b = 78 J
