For a certain transverse standing wave on a long string, an antinode is at x = 0 and an adjacent node is at x = 0.20 m. The displacement y(t) of the string particle at x = 0 is shown in the figure, where the scale of the y axis is set by ys = 4.3 cm. When t = 0.90 s, what is the displacement of the string particle at (a) x = 0.30 m and (b) x = 0.40 m ? What is the transverse velocity of the string particle at x = 0.30 m at (c) t = 0.90 s and (d) t = 1.3 s?
Answers
The expressions for the traveling and standing wave to find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
The traveling wave is a disturbance in the medium that moves at constant speed, in the case of a transverse wave the expression for the perpendicular oscillation is:
y = A sin (kx - wt)
Where y is the oscillation perpendicular to the direction of the displacement, A the amplitude, k in wave number and w the angular velocity.
Standing waves are formed when a traveling wave collides with an obstacle and is reflected, in this case the sum of the two waves gives a wave that does not shift in time and fulfills the relationship
[tex]\frac{\lambda}{2} = \frac{L}{n}[/tex]
Where λ is the wavelength, L the distance between the reflection points and n the number of nodes.
Indicates that for the standing wave the distance between an antinode and the node is x = 0.20 m, therefore
[tex]\frac{\lambda}{4} = \frac{L}{1}[/tex]
λ = 4L
λ = 4 0.20
λ = 0.80 m
The wave number.
k = [tex]\frac{2\pi }{\lambda }[/tex]
k = [tex]\frac{2 \pi }{0.80 }[/tex]
k = 2.5π i m⁻¹
In the associated traveling wave, from the graph we can see that the period of the wave is:
T = 2.8 s
the angular velocity is related to the period.
[tex]w=\frac{2\pi}{T} \\w = \frac{2\pi }{2.8}[/tex]
w = 0.714π rad/s
indicate the maximum displacement that is the amplitude of the wave.
A = [tex]y_s[/tex]
A = 4.3 cm
Let's write the equation of the traveling wave.
y = 4.3 sin [π (2.5 x - 0.714 t)]
with this expression we can answer the questions.
a) the displacement of the particle for x = 0.30 m
y = 4.3 sin (π (2.5 0.30 - 0.714 t))
y = 4.3 sin π( 0.75 - 0.714 t(
Remember that the angles must be in radians. For time t = 0 the displacement is
y = 4.3 0.707
y = 3.04 cm
b) The displacement for x = 0.4m
y = 4.3 sin (π 2.5 0.4)
y = 0 cm
c) the transverse velocity of the wave at x = 0.30 m for the time of t = 0.90s
the speed of the wave is
[tex]v= \frac{dy}{dt} \\v= A w cos ( kx - wt)[/tex]
v = 4.3 0.714π cos π(2.5 0.3 - 0.714 t)
v = 9.65 cos π(0.75 - 0.714 t)
For time t = 0.90 s the velocity is:
v = 9.65 cos π(0.75 - 0.714 0.9)
v = 9.65 0.9436
v = 9.11 cm / s
d) The velocity for time t = 1.3 s
v = 9.65 cos π(0.75 - 0.714 1.3)
v = 9.65 0.9999
v = 9.65 cm / s
In conclusion, using the expressions for the traveling and standing wave, we can find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
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A Vector that starts from
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49. A particle starts from rest at time t=0 and movies along the x axis. if the net force on is proportional to t its kinetic energy is proportional to?
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Answer:
F net ∞ [tex]\frac{1}{\sqrt{t} }[/tex]
Explanation:
In pic
_________________
(hopet his helps can I pls have brainlist (crown)☺️)
Why should people pay attention to scientists when making decisions?
A. Scientists are always correct.
B. Sclentists are the smartest people.
C. Scientists make the best decisions.
D. Scientists back up their statements with facts.
Answers
Answer:
D. Scientist back up their statements with facts.
Explanation:
The majority of the laws, theorems and thesis are base on with experiments. The follow different strategies for confirm the hypothesis with facts, one of the most famous models is the scientific method. That is a way of proof something in a trustworthy.
In the image is the resume about this method (graphical organizer).
what is a capacitor and capacitance.
Answers
Answer:
Capacitor is a device which used to store energy .
Capacitance is the ratio of the change in an electric charge in a system to the corresponding change in its electric potential.
Explanation:
A 0.015 kg marble sliding to the right at 0.225 m/s on a frictionless surface makes an elastic head-on collision with a 0.015 kg marble moving to the left at 0.180 m/s. After the collision, the first marble moves to the left at 0.180 m/s. Find the velocity of the second marble after the collision.
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Explanation:
[tex]if \: the \: masses \: of \: the \:two \: marbles\: equal \\ then \: each \: velocity \: sharing \: with \: other \: velocity[/tex]
In the popular game "Angry Birds" (refer Figure), for the red bird to hit the green bird with their range 55.0 m, what should be the launch angle (with respect to the horizontal) of the slingshot if the time of flight is 2.50 s?
Answers
Answer:
Explanation:
What are we to use as gravity in this imaginary world? I will ASSUME 9.81 m/s² and assume no air resistance.
horizontal velocity component
vx = 55.0/2.50 = 22 m/s
vertical initial velocity component
(12.0 - 10.0) = 0 + vy₀(2.50) + ½(-9.81)2.50²
vy₀ = 13.0625 m/s
θ = arctan(vy₀/vx) = arctan(13.0625/22) = 30.6997225... = 30.7°
The less dense areas created as a sound wave propagates are called
A. rarefactions✅
B. troughs
C. crests
D. compressions
Answers
Answer:
C. crests
don't trust me though
Answer:
The less dense areas created as a sound wave propagates are called rarefactions.
Explanation:
Hoped this helped.
Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale? What is the reading on the right scale?
Answers
The equilibrium conditions allow to find the results for the balance forces are:
F₁ = 225.4 N F₂ = 558.6 N
When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.
∑ F = 0
∑ τ = 0
Where F are the forces and τ the torques.
The torque is the product of the force and the perpendicular distance to the point of support,
The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.
We write the translational equilibrium condition.
F₁ - W₁ - W₂ + F₂ = 0
We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.
F₂ 2 - W₁ 1 - W₂ 1.5 = 0[tex]\frac{W_1 \ 1 + W_2 \ 1.5}{2}[/tex]
Let's calculate F₂
F₂ = [tex]\frac{W_1 \ 1 + W_2 \ 1.5 }{2}[/tex]
F₂ = (m g + M g 1.5)/ 2
F₂ = [tex]\frac{(12 + 68 \ 1.5 ) \ 9.8}{2}[/tex]
F₂ = 558.6 N
We substitute in the translational equilibrium equation.
F₁ = W₁ + W₂ - F₂
F₁ = (m + M) g - F₂
F₁ = (12 +68) 9.8 - 558.6
F₁ = 225.4 N
In conclusion using the equilibrium conditions we can find the forces of the balance are:
F₁ = 225.4 N F2 = 558.6 N
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Astone has a mass of 200 grams. When it is immersed in a measuring cylinder of water,the water rises 100 ml.What is the density of the stone
Answers
Answer:
2 g/mLExplanation:
The density of a substance can be found by using the formula
[tex]d = \frac{m}{v} \\ [/tex]
m is the mass
v is the volume
From the question
m = 200g
v = 100 mL
We have
[tex]d = \frac{200}{100} = 2 \\ [/tex]
We have the final answer as
2 g/mLHope this helps you
There are 270 students and teachers going on a field trip to a science center. If each school bus holds 54 people, how many buses are needed?
Answers
270/54 is five:)
While forming a 1.5kg aluminum statue, a metal smith heats the aluminum to 2700 degrees C, pours it into a mould, and then cools it to a room temperature of 23.0 degrees C. Calculate the thermal energy released by the aluminum during the process.
Answer is supposed to be: 4.7*10^6 j/kg?
Answers
The heat released by the aluminum during the process is [tex]3.6 \times 10^6 \ J[/tex].
The given parameters:
Mass of the statues, m = 1.5 kgFinal temperature of the status, t₂ = 2700 CTemperature when it is in the mould, t₁ = 23 ⁰CSpecific heat capacity of aluminum, C = 900The heat released by the aluminum during the process is calculated as follows;
[tex]Q = mc \Delta t \\\\Q = 1.5\times 900 \times (2700 - 23)\\\\Q = 3.6 \times 10^6 \ J[/tex]
Thus, the heat released by the aluminum during the process is [tex]3.6 \times 10^6 \ J[/tex].
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Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship
Answers
The length of the ship in terms of Emily's equal steps is 84.
The given parameters;
equal steps forward = 210equal steps backward = 42Defined parameters:
Let the constant velocity of the ship = VLet the velocity of Emily = VsLet the length of the ship = dLet the time of motion, = tThe velocity of Emily when moving forward in the direction of the ship:
[tex]V_s_1 = \frac{210}{t}[/tex]
The velocity of Emily when moving in opposite direction to the ship:
[tex]V_s_2 = \frac{42}{t}[/tex]
The constant velocity of the ship:
[tex]V = \frac{d}{t}[/tex]
Apply relative velocity formula to determine the length of the ship:
For forward (same direction) motion:
[tex](V_s_1 - V)t = d[/tex]
For backward (opposite direction) motion:
[tex](V_s_2 + V)t = d[/tex]
[tex](V_s_1 - V)t = (V_s_2 + V)t\\\\V_s_1 - V = V_s_2 + V\\\\\frac{210}{t} - \frac{d}{t} = \frac{42}{t} + \frac{d}{t} \\\\\frac{210}{t} - \frac{42}{t} = \frac{d}{t} + \frac{d}{t} \\\\\frac{210 - 42}{t} = \frac{d+ d}{t} \\\\\frac{168}{t} = \frac{2d}{t} \\\\2d = 168\\\\d = \frac{168}{2} \\\\d = 84[/tex]
Thus, we can conclude that the length of the ship in terms of Emily's equal steps is 84.
"Your question is not complete, it seems to be missing the following information;"
Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts 210 equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts 42 steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship
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groundwater returning to earths surface comes up through a ___?
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Answer:
Groundwater Returning to earths Surface come Up Though A drought
Explanation:
THIS DOES NOT NEED TO BE EXPLAINED IT'S EZZZ
A boat is headed north at a velocity of 8 km h-1. A strong wind is blowing whose pressure on the boat’s superstructure causes it to move sideways to the west at a velocity of 2 km h-1. There is also wave present that flows in a direction 30° south of east at a velocity of 5 km h-1. What is the velocity of the boat?
Answers
The velocity of the boat moving at the given conditions is 5.97 km/h at 67⁰.
The given parameters;
velocity of boat northwards, = 8 km/hvelocity of the boat westwards, = 2 km/hvelocity of the wind, = 5 km/h 30° south of eastThe resultant vertical velocity of the boat is calculated as;
[tex]v_y = 8 \ - \ 5 sin(30)\\\\v_y = 8 - 2.5\\\\v_y = 5.5 \ km/h[/tex]
The resultant horizontal velocity of the boat is calculated as;
[tex]v_x = -2 \ + 5cos(30)\\\\v_x = 2.33 \ km/h[/tex]
The resultant boat velocity is calculated as follows;
[tex]v = \sqrt{v_x^2 + v_y^2} \\\\v = \sqrt{2.33^2 + 5.5^2} \\\\v = 5.97 \ km/h[/tex]
The direction of the velocity;
[tex]\theta = tan^{-1} (\frac{v_y}{v_x} )\\\\\theta = tan^{-1}(\frac{5.5}{2.33} )\\\\\theta = 67 \ ^0[/tex]
Thus, the velocity of the boat moving at the given conditions is 5.97 km/h at 67⁰.
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Which is a force that wears away landforms? Select three options.
A. weathering
B. erosion
C. humans
D. clouds
E. light
Answers
i need help with these please
Answers
Answer:
Explanation:
A mass is placed at the end of a spring. It has starting velocity of V & allowed to oscillate freely. If the mass has a starting velocity of 2V, what would the period be? Half as long, remains constant, 1/4 as long , 4 times as long, or 2 times as long?
Answers
Answer:
Equation for SHM can be written
V = w A cos w t where w is the angular frequency and the velocity is a maximum at t = 0
V1 = w1 A cos w1 t
V2 = w2 A cos w2 t
V2 / V1 = w2 / w1 since cos X t = 1 if t = zero
V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2
If the velocity is twice as large the period will be 1/2 long
At a distance of 1 cm from the source, the flux from the isotropic source is one hundred times brighter. Which source is brighter fifty centimeters away
Answers
Answer:
op
Explanation:
At a distance of 1 cm from the source, the flux from the isotropic source is one hundred times brighter. Which source is brighter fifty centimeters away
Explain different layers of atmosphere and the pressure in each layer. Draw diagram
Answers
Answer:
Our atmosphere has five different layers. They are:
1. Troposphere: This is the most important layer of the atmosphere with an average height of 13 km from the earth. It is in this layer that we find the air that we breathe. Almost all the weather phenomena such as rainfall, fog and hailstorm occur here.
2. Stratosphere: This layer extends up to a height of 50 km. It presents the most ideal condition for flying airplanes. It contains a layer of ozone gas which protects us from the harmful effect of the sun rays.
3. Mesosphere: This layer extends up to a height of 80 km. Meteorites bum up in this layer on entering from the space.
4. Thermosphere: In this layer, the temperature rises very rapidly with increasing height. The ionosphere is a part of this layer. It extends between 80-400 km. This layer helps in radio transmission. Radio waves transmitted from the earth the reflected back to the earth by this layer.
5. Exosphere: It is the uppermost layer where there is very thin air. Light gases such as helium and hydrogen float into space from here.
An archer's bow is drawn at its midpoint until the tension in the string is 0.842 times the force exerted by the archer. What is the angle between the two halves of the string
Answers
Consult the attached free body diagram.
If we take the direction of F to be the positive horizontal axis, and upward to be the positive vertical axis, then using Newton's second law we have net forces
• ∑ F [horizontal] = F [archer] + T cos(180° - θ) + T cos(180° + θ) = 0
• ∑ F [vertical] = T sin(180° - θ) + T sin(180° + θ) = 0
since the bow is held in place while it's drawn. T is the magnitude of the tension in the string, and it can be shown to be equal in both strings since they both make the same angle with the negative horizontal axis (the dashed line).
We only really need the first equation. Simplifying it, we get
F [archer] - T cos(θ) - T cos(θ) = 0
F [archer] - 2T cos(θ) = 0
F [archer] = 2T cos(θ)
cos(θ) = F [archer] / (2T)
We're given that the tension T in the string is 0.842 times the force exerted by the archer, which is to say
T = 0.842 F [archer]
and from this we have
cos(θ) = F [archer] / (2 • 0.842 F [archer])
cos(θ) = 1/1.684
cos(θ) ≈ 0.593
Solving for θ gives an angle of θ ≈ arccos(0.593) ≈ 53.6°. Then the angle between the two tension forces is twice this, or about 2θ ≈ 107°.
Which light is most sensitive to the eyes?
Answers
Answer:
Our eyes are most sensitive to the wavelengths corresponding to the yellow and green colors of the spectrum. Flashy signs and some fire engines are painted in a yellowish-green color to attract our attention.
2. An auditorium has 58 seats in the first row, 62 seats in the second row, 66 seats in the third row, and so
on.
a)Find the explicit formula of this arithmetic sequence.
B) find the number of seats in the twentieth row.
Answers
Two identical vertical springs S1 and S2 have masses m1 = 400 g and m2 = 800 g attached to them. If m1 causes spring S1 to stretch by 4 cm, what is the ratio of the potential energy of S1 and S2? Use g = 10 m/s^2
Select one:
a. 1:2
b. 4:1
c. 1:4
d. 1:3
e. 2:1
Answers
Answer:
potential energy = mgh
= 400÷1000 × 10× 4÷100
= 0.4 × 10 × 0.04
=4/10 ×10×4/100
= 4/10 × 4/10
=16/100
= 0.16 joules
m1 (400) stretches 4cm
m1 (100g) stretches 1cm
so, m2(800g) stretches 8 cm
potential energy of m2 = mgh
= 800/1000 ×10×8/100
= 0.8 × 0.8
=8/10 ×8/10
= 64/100
=0.64 joules
Ratio of s1 to s2
16/100 ÷ 64/100
= 1:4 ( answer)
Describe all the ways a bicyclist can accelerate
Answers
the momentum of a car before the crash is 22500kg m/s. the car stops in 0.14s. what is the average force on the car during the crash?
Answers
Answer:
what is the full questio n because I can't see the momentum
Help please..
Kepler’s third law states that:
A. the orbits of the planets are elliptical.
B. the planets move slower when they are closer to the Sun and faster when they are farther from the Sun.
C. the square of the ratio of the periods of any two planets revolving around the Sun is equal to the cube of the ratio of their average distance from the Sun.
D. objects attract other objects with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them.
Answers
Answer:
c
Explanation:
6. A 15.53 kg bag of soil falls 5.50 m at a construction site. If all the energy is retained by the soil in the bag, how much will its temperature increase
Answers
The increase in the temperature of the soil when bag fall from the given height is 0.065 ⁰C.
The given parameters:
mass of the bag, m = 15.53 kgheight of fall, h = 5.5 mspecific heat capacity of soil, C = 0.200 kcal/kg ⁰CApply the principle of conservation of energy as follows;
[tex]mgh = mC \Delta T\\\\gh = C \Delta T\\\\\Delta T= \frac{gh}{C}[/tex]
Convert the value of C in kcal/kg ⁰C to J/kg ⁰C
1 kcal = 4180 J
[tex]0.200 \ kcal/kg ^0 C= 0.2 \times 4180 (J/kg ^0C) = 836 \ J/kg^0 C[/tex]
The increase in the temperature of the soil is calculated as follows;
[tex]\Delta T= \frac{gh}{C}\\\\\Delta T= \frac{9.8 \times 5.5}{836} \\\\\Delta T= 0.065 \ ^0C[/tex]
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