Answer:
a1 = -w^2 A sin w t where w is the angular frequency
a2 = - (2 w)^2 A sin 2 w t where w2 = 2 w1
a2 / a1 = 4 sin 2 w t / sin w t
Since sin 2 w t / sin w t = 2 sin w t will change twice as fast
a2 / a1 = 8 the max acceleration must change 8 tims as fast
In an earlier chapter you calculated the stiffness of the interatomic "spring" (chemical bond) between atoms in a block of lead to be 5 N/m. Since in our model each atom is connected to two springs, each half the length of the interatomic bond, the effective "interatomic spring stiffness" for an oscillator is 4*5 N/m = 20 N/m. The mass of one mole of lead is 207 grams (0.207 kilograms).
What is the energy, in joules, of one quantum of energy for an atomic oscillator in a block of lead?
Answer:
8.01e-22
Explanation:
The energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.
Extension produced by one mole of leadThe extension produced by one mole of lead atom is calculated by applying Hooke's law;
F = kx
mg = kx
x = mg/k
x = (0.207 x 9.8) / (20)
x = 0.101 m
Energy stored in the lead blockThe Energy of one quantum of energy for an atomic oscillator in a block of lead is calculated as follows;
E = ¹/₂kx²
E = ¹/₂ (20)(0.101)²
E = 0.102 J
Thus, the energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.
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please help asap 100 POINTS and BRAINLIEST to best answer.
D. :)
lmk if im right or wrong
Answer:
D. 5ºC
Explanation:
Not B, because 1k= -457.87 fahrenheit. That means it is also not C. Since the icicles are melting, then that means it is not A either.
5ºC=about 41ºF
Una turbina de vapor
recibe vapor con un flujo másico de 30 kg/s a 6205 kPa, 811 K, con una velocidad a la
entrara de 10 m/s. El vapor a la entrada tiene una energía interna específica de 3150.3
kJ/kg y un volumen específico de 0.05789 m3
/kg. El vapor sale de la turbina a 9.859 kPa,
318.8 K. El vapor sale a 200 m/s con una energía interna específica de 2211.8 kJ/kg y
un volumen específico de 13.36 m3
/kg. Encuentre la potencia producida por la turbina
si ésta pierde calor a una tasa de 30 kW.
Este problema está describiendo una turbina de vapor a la que entra vapor a 30 kg/s, 6.205 kPa y 811 K con una velocidad de 10 m/s y sale a 9.859 kPa, 318.8 K y con una velocidad de 200 m/s. Adicionalmente, tanto el volumen específico como la energía interna son dados para ambas corrientes.
Con lo anterior, resulta posible escribir un balance de energía para esta turbina, despreciando todo efecto por energía potencial ya que no hay diferencia significativa entre la altura de la entrada (1) y la salida (2), pues están practicamente al mismo nivel:
[tex]mh_1+\frac{1}{2} mv^2_1=mh_2+\frac{1}{2} mv^2_2+Q_2+W_2[/tex]
Aquí vemos que la incógnita es [tex]W_2[/tex] como la potencia que produce la turbina. Ahora, el primer cáculo a realizar es el de las entalpías de las corrientes de entrada y salida, dada la energía interna, presión y volumen específico:
[tex]h_1=3150.3\frac{kJ}{kg}+6205kPa*0.05789\frac{m^3}{kg} =3509.51\frac{kJ}{kg}\\\\h_2=2211.8\frac{kJ}{kg}+9.859kPa*13.36\frac{m^3}{kg} =2342.72\frac{kJ}{kg}[/tex]
Ahora, podemos reacomodar el balance de energía con el fin de resolver [tex]W_2[/tex]:
[tex]W_2=m(h_1-h_2)+\frac{1}{2} m(v^2_1-v^2_2)-Q_2[/tex]
Finalmente, reemplazamos los valores para obtener:
[tex]W_2=10\frac{kg}{s} (3509.51-2342.72)\frac{kJ}{kg} +\frac{1}{2} *10\frac{kg}{s} [(10\frac{m}{s}) ^2-(200\frac{m}{s} )^2]*\frac{1kJ}{1000J} -30\frac{kJ}{s}\\\\W_2=11438.4 kJ/s=11438.4kW[/tex]
Es de precisar que la energía cinética como 1/2 m*v² resulta en Joules, por lo que hay que convertir a kilojoules para tener unidades consistentes de kilowatts al final.
Aprende más:
https://brainly.com/question/21902769https://brainly.com/question/24322350Comparing energy resources
Answer:
Sorry, This Photo is not clear.
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
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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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.
Answer:
c
Explanation:
A 45 kg figure skater is spinning on the toes of her skates at 1.1 rev/s. Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 kg, 20 cm average diameter, 160 cm tall) plus two rod-like arms (2.5 kg each, 71 cm long) attached to the outside of the torso. The skater then raises her arms straight above her head. In this latter orientation, she can be modeled as a 45 kg, 20-cm-diameter, 200-cm-tall cylinder.
What is her new rotation frequency, in revolutions per second?
Answer: HOPE THIS HLEP YOU
Explanation:
Carrie is driving a car. Which factors determine the kinetic energy of her car
at this point
Answer:
The two main factors that affect kinetic energy are mass and speed.
Explanation:
Kinetic energy is the energy that is caused by the motion. The kinetic energy of an object is the energy or force that the object has due to its motion. Your moving vehicle has kinetic energy; as you increase your vehicle's speed, your vehicle's kinetic energy increases.
Have a great day! :D
I will mark brainlist
If a wave’s amplitude is 2cm, then its height is equal to:
5 cm.
0 cm
4 cm.
2 cm
Answer:
4 cm
Explanation:
Amplitude is the measure from the MIDLINE to the peak of the wave....so the wave HEIGHT is twice the ampltude
2 x 2 cm = 4 cm
according to Newton's __ law, an object with no net force acting on it remains at rest or in motion with a constant veloctiy
Answer:
i think its his law of inertia
Explanation:
this law is about motion
Using the expression for the total energy of this system, it is possible to show that after the switch is closed, d2qdt2=−kq, where k is a constant. Find the value of the constant k.
The value of the constant K is [tex]\mathbf{K = \dfrac{1}{LC}}[/tex]
According to Kirchhoff's loop rule, the total algebraic sum of potential differences in any loop, combining voltage provided by voltage sources as well as resistive components, must equal zero.
Thus, the relation for Kirchhoff's loop rule can be expressed as:
[tex]\mathbf{\dfrac{q}{c}- L\dfrac{dI}{dt} = 0}[/tex]
We all know that the current in the nonconstant charge flow can be written as:
[tex]\mathbf{I = \dfrac{dq}{dt}}[/tex]
∴
Replacing the current (I) into Kirchhoff's loop rule, we have:
[tex]\mathbf{ L\dfrac{d}{dt} ( \dfrac{dq}{dt})= -\dfrac{q}{c}}[/tex]
[tex]\mathbf{ \dfrac{d^2q}{dt^2}= -\dfrac{q}{Lc} \ \ ---(1)}[/tex]
From the given question, when the switch is closed
[tex]\mathbf{ \dfrac{d^2q}{dt^2}= -kq\ \ ---(2)}[/tex]
Then, the charges on the capacitor start to b, resulting in the rise of the current in the circuit.
∴
By equating both equations (1) and (2);
[tex]\mathbf{K = \dfrac{1}{LC}}[/tex]
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In which of the following situations is the Doppler Effect absent?
The source and the observer are moving towards each other.
The observer is moving toward the source.
The source is moving away from the observer.
The source and the observer are both moving with the same velocity.
Answer:
Correct option is
C
The listener is moving towards the source of sound
The Doppler effect (or the Doppler shift) is the change in frequency or wavelength of a wave (or another periodic event) for an observer moving relative to its source.
So if both listener and source of sound are stationary or are moving with the same velocity that is relative velocity is zero, there won't be any change in frequency.
Thus, in this case, the effect will arise when a listener is moving towards the source of the sound.
PLEASE HELP BRAINLIEST TO THE RIGHT ANSWER!!!!!
A 7.8 kg object is suspended by a string from the ceiling of an elevator. The acceleration of gravity is 9.8 m/s^2.
A. Determine the tension in the string if it is accelerating upward at a rate of 1.5 m/s^2. Answer in units of N.
B. Determine the tension in the string if it is accelerating downward at a rate of
1.5 m/s^2
Gravitational pull downward on the object: 7.8 x 9.8 = 76.44N
A. 7.8 x 1.5 = 11.7N upward force
Tension = 76.44 + 11.7 = 88.14 N
Answer: 88.14 N
B. 76.44 - 11.7 = 64.74N
Answer: 64.74 N
Answer:
88.14 N
64.74 N
Explanation:
HIHIHHHHHHHHIHIHIIHIHHHHIHIHIHIHIHIHIHIHIHIHIHIHIHIIH
Answer:
Ok thanks for the points though
Explanation:
A block slides down a smooth ramp, starting from rest at a height h. When it reaches the bottom it's moving at speed v. It then continues to slide up a second smooth ramp. At what height is its speed equal to v/2
Answer:
3h/4
Explanation:
At speed v/2 height will be 3/4 h
What is equation of motion in kinematics?Equation that describes the motion of point , bodies , and system of bodies without considering the force that cause them to move is called equation of motion in kinematics
When block is at top of first ramp
u=0 ( block was at rest )
a = g ( acceleration due to gravity
using equation of motion
2as = v^2 - u^2
2gh = v^2
Then the block continued and reached a speed of v1 = v/2 on second ramp
now , final velocity = v= v1 =[tex]\sqrt{2gh}[/tex] / 2
u= [tex]\sqrt{2gh\\}[/tex]
s= h1
using equation of motion , we get
2as = v^2 - u^2
2(-g)h1 =( [tex]\sqrt{2gh}[/tex]/2)^2 - [tex]\sqrt{2gh}[/tex]
2(-g)h1 = (g h - 4 g h) / 2
h1 = 3/4 h
At speed v/2 height will be 3/4 h
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Question 3 of 20 :
Select the best answer for the question.
3. Which of the following accurately describes the behavior of water when subjected to
temperature change?
A. The volume of water will decrease if heated from 6°C to 7°C.
B. The volume of water will increase if cooled from 3°C to 2°C.
O C. A mass of water will contract if cooled from 1°C to 0°C.
D. A mass of water will expand if heated from 0°C to 2°C.
As water cools the volume expands.
Answer: B. The volume of water will increase if cooled from 3°C to 2°C.
Which is the best way to avoid plagiarism? O use more than one source from your research O repeat information from your sources O present a topic in your own words O understand your topic before you begin writing
Answer:
O present a topic in your own words
Explanation:
Plagiarism is using someone else's words or work without giving them proper credit. Presenting a topic in your own words means you are not copying anyone.
Answer:
present a topic in your own words
Explanation:
A phone is dropped from the roof of the school building. The building is 13.2 meters high. How long was the phone in the air before it crashed on the ground?
Answer:
5scs
Explanation:
phones fall fast so i guessed
An empty plastic or glass dish being removed from a microwave oven is cool to the touch. How can this be possible
What is the possible resultant of a 2-unit vector and 4-unit vector?
Answer:
Explanation:
resultant could have magnitude 4 - 2 ≤ V ≤ 4 + 2 or 2 ≤ V ≤ 6
If θ is the angle of the 4 unit vector, the resultant could have angle φ in the range of
φ = θ ± arcsin (2/4)
θ - 30° ≤ φ ≤ θ + 30°
The possible resultant of a 2-unit vector and 4-unit vector is 6 and 2.
What is a Unit vector?The quantity which has both the magnitude and direction is known as vector.The Examples for Vector quantity is Acceleration, velocity, force and displacement, etc.,.The quantity which has only the magnitude but has no direction, (examples include speed, time and distance) can be known as scalar.If the magnitude of vector is 1, then it can be the Unit vector.
In the Unit vector, x axis will have the vector of i, y axis will have the vector j, z axis will have the vector k.
When two vectors are in same direction then we add up to get their resultant and in this case resultant would be maximum
R = A + B
R = 4 + 2
R = 6
If vectors are in opposite direction then they provide minimum resultant and in this case we subtract them
R = A - B
R = 4 - 2
R = 2
Thus, the possible resultant are 6 and 2.
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At a construction site, a 68.0 kg bucket of concrete hangs from a light (but strong) cable that passes over a light friction-free pulley and is connected to an 85.0 kg box on a horizontal roof (see the figure (Figure 1)). The cable pulls horizontally on the box, and a 46.0 kg bag of gravel rests on top of the box. The coefficients of friction between the box and roof are shown. The system is not moving.
Following are the calculation to the friction force:
Please see the diagram below for an illustration of the forces acting on objects.Therefore, m denotes the mass of a gravel bag, and M denotes the overall mass of the box.Whenever the system is stationary, the friction force on the box equals the tension in the string.The substring tension force is provided as follows:
[tex]\to T-m_{1}g=0 \\\\\to T = m_{1}g \\\\[/tex]
[tex]= (68\ kg) (9.81 \ \frac{m}{s^2} ) (\frac{ 1\ N}{ 1\ kg \cdot \frac{m}{s^2}}) \\\\ = (68 ) (9.81 ) ( 1\ N) \\\\= 667.08\ N \\\\[/tex]
Therefore, the friction force on the box is "667 N".
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Ignoring any effects of dc resistance, what is the total reactance of a 250 mH coil in series with a 4.7 microfarad capacitor at a signal frequency of 60 Hz
The total reactance of the inductor and the capacitor is 470.1 ohms.
The given parameters;
inductance of the coil, L = 250 mHcapacitance, C = 4.7 μfrequency of the circuit, f = 60 HzThe inductive reactance of the coil is calculated as follows;
[tex]X_l = \omega L\\\\X_l = 2\pi f L\\\\X_l = 2\pi \times 60 \times 250 \times 10^{-3}\\\\X_l = 94.26 \ ohms[/tex]
The capacitive reactance of the capacitor is calculated as follows;
[tex]X_c = \frac{1}{\omega C} \\\\X_c = \frac{1}{2\pi f C} \\\\X_c = \frac{1}{2\pi \times 60 \times 4.7 \times 10^{-6}} \\\\X_c = 564.31 \ ohms[/tex]
The impedance of the circuit is calculated as follows;
[tex]Z = \sqrt{(X_c - X_l)^2} \\\\Z = X_c - X_l\\\\Z = 564.31 - 94.26\\\\Z = 470.1 \ ohms[/tex]
Thus, the total reactance of the inductor and the capacitor is 470.1 ohms.
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A car driving on level ground at 20.0 m/s slams on its brakes and skids to a halt. If the coefficient of kinetic friction between the car’s tires and the road is 0.750, how far does the car skid before stopping? How far would the car have skidded if it had been moving at 40.0 m/s?
The distance the car has skidded if it had been moving at 40.0 m/s is 27.2m
The linear force acting on the car is opposed by the frictional force. Hence;
[tex]F=F_f[/tex]
[tex]ma = -\mu R\\ma =-\mu mg[/tex]
m is the mass of the car
a is the acceleration
[tex]\mu[/tex] is the coefficient of friction
R is the normal force
Given the following parameters
[tex]a=-\mu g[/tex]
[tex]a =-0.75(9.8)\\a=-7.35m/s^2[/tex]
Get the distance the car has skidded if it had been moving at 40.0 m/s
[tex]v^2=u^2+2as[/tex]
[tex]0^2=20^2+2(-7.35)s\\0=400-14.7s\\s=\frac{400}{14.7}\\s= 27.2m[/tex]
Hence the distance the car has skidded if it had been moving at 40.0 m/s is 27.2m
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An ambulance is currently traveling at 25 m/s, and is accelerating with a constant acceleration of 5 m/s^2. The ambulance is attempting to pass a car which is moving at a constant velocity of 50 m/s. How far must the ambulance travel until it matches the car’s velocity?
Answer:
PLEASE MARK ME BRAINLIEST!!!
Explanation:
Given initial velocity of ambulance
v0=18m/s , acceleration a=5m/s2
To find distance, x=?
First we need to calculate the time for which it acquires 30\,m/s. For that use equation
v=v0+at
30=18+5×t
⇒t=30−185=125seconds
Distance travelled by the ambulance
x=v0t+12at2
x=(18×125)+12×5×(125)2
x=43.2+2.5×5.76=43.2+14.4
x=57.6m
Therefore the ambulance has to travel 57.6 m to match the velocity of car.
Your in an escape room. Only its real and your oxygen runs out in 3 minutes unless you solve this problem, The problem :a ball is thrown and travels 30 inches before bouncing, It bounces and travels 50% of the distance traveled. It continues to do this until coming to rest. What is the total distance traveled by ball? Also it's cold and dark and you have no phone nor anyone to ask. You have no ball
60 inches
because if 30=50% then 60=100% so its 60 inches
The sum of the series allows to find the result for the total distance that the ball bounces is:
total distance = 59.52 in
A series is a set of things or numbers related by a specific operation.
They indicate that the ball falls from an initial height y₀ = 30 in. and it bounces 50% of the height and the process is repeated until it stops, see attached.
Let's build a table to observe the sequence.
drop height rebound
1 30 15
2 15 7.5
3 7.5 3.75
If we call the first term y₀
The first bounce can be found.
y₁ = [tex]\frac{y_o}{2}[/tex]
The second bounces.
[tex]y_2 = \frac{y_1}{2} \\y_2 = \frac{y_o}{4}[/tex]
The third bounce.
[tex]y_3 = \frac{y_2}{2} \\y_3 = \frac{y_0}{ 8}[/tex]
By observing this table we can construct a series of the form
Total distance = [tex]y_o \ ( 1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8} + ... +\frac{1}{2n} )[/tex]
The sum of the serie has a result of
sum = 127/64 = 1,984
Let's calculate
distance total = 30 1,984
Distance total = 59.52 cm
In conclusion, using the sum of the series we can find the result for the total distance that the ball bounces is:
total distance = 59.52 in
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A 27 kg chair initially at rest on a horizontal
floor requires a 173 N horizontal force to set
it in motion. Once the chair is in motion, a
148 N horizontal force keeps it moving at a
constant velocity.
The acceleration of gravity is 9.81 m/s.
a) What is the coefficient of static friction
between the chair and the floor?
Answer:
4653
4565445655687568677667876
The speed of light inside a medium is (2.0 x 10^8m/s) What is the index of refraction (n) of the medium?
Answer:
Refractive index of a medium = Speed of light in vacuum/Speed of light in a medium
1.5 = 3 x 108 / Speed of light in medium
Speed of light in the medium = 3 x 108 /1.5
= 2 x 108 m/s.
Explanation:
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?
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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5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal while you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.
Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.
By Newton's second law, the net vertical force is
• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0
where a is the acceleration of the wagon.
Solve for n (the magnitude of the normal force) :
n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N
Then
f = 0.500 (58.0 N) = 29.0 N
Meanwhile, the horizontal component of the applied force has magnitude
(80.0 N) cos(30.0°) ≈ 69.3 N
Now calculate the work done by either force.
• friction: -(29.0 N) (10.0 m) = -290. J
• pull: (69.3 N) (10.0 m) = 693 J
Harriet is practicing archery. As she draws back an arrow, the arrow has LaTeX: 34.3 J of elastic potnetial energy, and LaTeX: 0.40 J of gravitational potential energy. How much kinetic energy will the arrow have when it is about to hit the ground?
The kinetic energy of the arrow when it is about to hit the ground is 34.7 J.
The given parameters;
elastic potential energy of the arrow, Ux = 34.3 Jgravitational potential energy, P.E = 0.4 JAccording the principle of conservation of energy, the total potential energy of the arrow will be converted to maximum kinetic energy when the arrow is about to hit the ground.
[tex]K.E = P.E + U_x\\\\K.E = 34.3 \ J \ + \ 0.4 \ J\\\\K.E = 34.7 \ J[/tex]
Thus, the kinetic energy of the arrow when it is about to hit the ground is 34.7 J.
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