f = force = 5N
v = velocity = 4m/s
p = power
p = force * velocity
Replacing:
p= 3N * 4 m/s = 12W
The motor produces 12W of power.
Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00 ✕ 105 kg on springs that have adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven—the driving force is transferred to the object, which oscillates instead of the entire building.(a) What effective force constant (in N/m) should the springs have to make the object oscillate with a period of 2.10 s? N/m(b) What energy (in J) is stored in the springs for a 1.80 m displacement from equilibrium? J
ANSWERS
(a) 3.58 x 10⁶ N/m
(b) 5.8 x 10⁶ J
EXPLANATION
Given:
• The mass of the object hanging from the spring, m = 4.00 x 10⁵ kg
Find:
• (a), The force constant the springs should have to make the object oscillate with a period of 2.10 s
,• (b), The energy stored in the springs for a 1.80 m displacement from equilibrium.
(a) The period of an object in a spring is,
[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex]Where m is the mass of the object and k is the spring constant.
We want to find k for T = 2.10 s, so we have to solve this equation for k, which gives us,
[tex]k=m\cdot\left(\frac{2\pi}{T}\right)^2[/tex]Replace the known values and solve,
[tex]k=4\cdot10^5\cdot\frac{4\pi^2}{(2.1)^2}\approx3,580,808.85N/m\approx3.58\cdot10^^6N/m[/tex]Hence, the force constant is 3.58 x 10⁶ N/m.
(b) The energy stored in a spring with force constant k when the mass is displaced x meters from equilibrium is,
[tex]U_s=\frac{1}{2}kx^2[/tex]Replace the known values and solve to find the energy stored in this spring,
[tex]U_s=\frac{1}{2}\cdot3.58\cdot10^6N/m\cdot1.8^2m^2\approx5.8\cdot10^6J[/tex]Hence, the energy stored in the spring for a 1.80 m displacement from equilibrium is 5.8 x 10⁶ J.
Two 4.137 cm by 4.137 cm plates that form a parallel-plate capacitor are charged to +/- 0.531 nC.What is the electric field strength inside the capacitor if the spacing between the plates is 2.198 mm?
The electric field of two parallel plates is given by:
[tex]\begin{gathered} E\cdot d=\frac{Q}{d\cdot A\epsilon o} \\ so: \\ E=\frac{Q}{A\epsilon o} \end{gathered}[/tex]Where:
[tex]\begin{gathered} \epsilon o=8.8542\times10^{-12} \\ A=0.04137^2=1.71\times10^{-3}m^2 \\ Q=0.531\times10^{-9}C \end{gathered}[/tex]Therefore:
[tex]\begin{gathered} E=\frac{0.531\times10^{-9}}{(8.8542\times10^{-12})(1.71\times10^{-3})} \\ E\approx35071V/m \end{gathered}[/tex]Which formula is equivalent to
v = m / D
v = mD / v
v = D x m
v = D / m
Answer:
i don't know.
Explanation: I = me Don't = Don't know = know
The potential difference across the ends of a wire is doubled in magnitude. If Ohm's law is obeyed, which one of the following statements concerning the resistance of the wire is true?a- The resistance is one half of its original value.b- The resistance decreases by a factor of four.c- The resistance is twice its original value.d- The resistance is not changed.e- The resistance increases by a factor of four.
Given:
The potential difference across the ends of a wire is doubled in magnitude
To find:
which one of the following statements concerning the resistance of the wire is true
Explanation:
The potential difference is doubled in magnitude. We know, according to Ohm's law,
[tex]V=IR[/tex]I is the current and R is the resistance.
If the potential difference increases, the product of I and R also increases.
Now the resistance depends as,
[tex]\begin{gathered} R=\frac{\rho l}{A} \\ \rho\text{ is the resistivity, l is the length and A is the area of the conductor.} \end{gathered}[/tex]The resistance is independent of potential differences. When the potential difference increases, the current also increases and the resistance remains the same.
Hence, The resistance is not changed.
How much work must you do to stretch the spring 0.030 m ? and how much work must you do to compress it 0.030m ?
Given:
The force constant of spring is
[tex]k\text{ =2.5}\times10^4\text{ N/m}[/tex]To find the work done when spring is stretched to 0.03 m and work done when compressed to 0.03 m.
Explanation:
The work done to stretch the spring will be
[tex]\begin{gathered} W\text{ =}\frac{1}{2}kx^2 \\ =\frac{1}{2}\times2.5\times10^4\times(0.03)^2 \\ =\text{ 11.25 J} \end{gathered}[/tex]The work done to compress the spring will be
[tex]\begin{gathered} W\text{ =-}\frac{1}{2}kx^2 \\ =-\frac{1}{2}\times2.5\times10^4\times(0.03)^2 \\ =-11.25\text{ J} \end{gathered}[/tex]A ball is thrown vertically upward with a speed of 36.0 m/s.(a) How high does it rise? m(b) How long does it take to reach its highest point? s(c) How long does the ball take to hit the ground after it reaches its highest point? s(d) What is its velocity when it returns to the level from which it started? m/s
ANSWER:
(a) 66.1 m
(b) 3.67 s
(c) 3.67 s
(d) -36 m/s
STEP-BY-STEP EXPLANATION:
Given:
Initial velocity (u) = 36 m/s
Final velocity (v) = 0 m/s
(a)
We can calculate the height using the following formula:
[tex]\begin{gathered} h=\frac{v^2-u^2}{2g} \\ \\ \text{ We replacing:} \\ \\ h=\frac{0^2-(36)^2}{(2)(-9.8)} \\ \\ h=66.1\text{ m} \end{gathered}[/tex](b)
Now, we calculate the time as follows:
[tex]\begin{gathered} v=u+gt \\ \\ t=\frac{v-u}{g} \\ \\ \text{ we replacing} \\ \\ t=\frac{0-36}{-9.8} \\ \\ t=3.67\text{ s} \end{gathered}[/tex](c)
The time it takes for the ball to hit the ground after reaching its highest point is the same as the time it takes to reach it, which means that it is equal to 3.67 seconds
(d)
The velocity it takes to return is also the same but being in the opposite direction, it has the same magnitude but with the opposite sign, that is, with a minus sign.
So the velocity when it returns to the level it started from is -36 m/s
Hi I just answered a question for electricity grade 12 could you please if it is correct
F = k*(q1)*(q2)/r^2
We know that:
F = 71 N
k = coulomb's constant = 9*10^9
q1 = ?
q2 = ?
We know that q1 = q2, since in the question it states that the charges are identical.
r = 0.99 m
F = k*(q1)*(q1)/r^2
F = k*(q1)^2/r^2
71 = 9*10^9 * (q1)^2 / 0.99^2
Isolating (q1)^2,
71 * 0.99^2 / (9*10^9) = (q1)^2
Taking square root of both sides to get q1 to the power of 1,
sqrt( 71 * 0.99^2 / (9*10^9) ) = q1
Now, simplifying the left side in a calculator, we get that
q1 = 8.793 * 10^-5 C = q2
Can you help me with this question? Find the sum of -2x^2 - 4x - 12 and - x^2 - 2x The answer I got was -3x^2 - 6x - 12 Just want to see if I did it correctly*Last tutor got disconnected
You have the following sum of expression:
(-2x^2 - 4x - 12) + (-x^2 - 2x)
In order to find the sum, simplify like terms (terms with the same variable and same exponent). Then, you have:
(-2x^2 - 4x - 12) + (-x^2 - 2x) =
-2x^2 - x^2 - 4x - 2x -12 =
-3x^2 - 6x - 12
Hence, the answer is -3x^2 - 6x - 12
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
5. Draw the ray diagram to show the image formed by a concave lens with focal length 5cm whenthe object is placed 15 cm from the lens. What is the magnification?
Given:
The focal length of the concave lens is f = -5 cm
The object distance from the lens is u = -15 cm
Required:
Ray diagram
Magnification
Explanation:
The rules related to incident ray and reflected ray are:
1) Any incident ray moving parallel to the principal axis will pass through focus after reflection.
2) Any incident ray passing through the focus will move parallel to the principal axis after reflection.
3) Incident ray passing through the optical center will reflect undeviated.
The ray diagram is shown below
Here, yellow lines represent incident rays passing parallel to the principal axis and the reflected ray passing through the focus.
The maroon line shows the incident ray passing through the optical center and the undeviated reflected ray.
The green color is the object and image.
In order to find magnification, first, we need to calculate the image distance given as
[tex]\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \\ =-\frac{1}{5}-\frac{1}{15} \\ v=-3.75\text{ cm} \end{gathered}[/tex]The magnification will be
[tex]\begin{gathered} m\text{ = }\frac{v}{u} \\ =\frac{-3.75}{-15} \\ =0.25\text{ } \end{gathered}[/tex]Final Answer: Thus, the magnification is 0.25 which means the size of the image is 0.25 times the object.
The volume of a cylinder with height h and radius r can be found using the formula V = πr2hFind the volume of a cylinder with radius 8 feet and height 7 feet.__ ft3
In order to calculate the volume of this cylinder, let's use the given values of radius and height in the formula.
Using π = 3.1416, we have:
[tex]\begin{gathered} V=\pi r^2h \\ V=3.1416\cdot8^2\cdot7 \\ V=1.407.44 \end{gathered}[/tex]So the volume is 1407.44 ft³.
You call out in a loud voice and hear an echo of your voice. Where are you most likely located?A.In a boat on a lakeB.In an open fieldC.In a canyonD.In a forest
We will have the following:
Since sound bounces off of surfaces particularly ones that are perpendicular, then in this scenario you are most likely located in a canyon.
The picture below shows a skateboarder just before he drops down a ramp. Assume the skateboarder does not experience any friction.Which correctly shows the potential energy and kinetic energy of the skateboarder as he moves down the left side of the ramp and up to the top of the right side?
We are given that a skater is on top of a ramp. At this point the skater has the maximum gravtitaional potential energy "PE" and a minimum kinetic energy "KE". The gravitational potential enery depends on the height of the skater and the kinetic energy depends on his velocity.
When the skater is at the bottom of the ramp his gravitational potential energy is minimum since his height is the lowest and his kientic energy is maximum since all the gravitational potential eerngy has turn into kinetic energy. This means that at the lowest point the velocity is also maximum.
Whe teh sketer reaches the highest point in the right side his kinetic energy has converted again into gravitational potential energy since the height has increased. Now, the kinetic energy is minimum and the gravitational potential energy is maximum.
This can be seen represented next:
The graph should be a representation of the fluctuations in kinetic and potential energy, therefore, the only graph that represetns this is B.
Object A, which has been charged to +13.13 nC, is at the origin. Object B, which has been charged to -26.7 nC, is at x=0 and y=2.72 cm. What is the magnitude of the electric force on object A?
ANSWER:
0.004259 N
STEP-BY-STEP EXPLANATION:
Given:
qA = +13.13 nC
qB = -26.7 nC
d = y - x = 2.72 - 0 = 2.72 cm = 0.0272 m
We use Coulomb's law to calculate the force:
[tex]F=k\cdot\frac{q_A\cdot q_B}{d^2}[/tex]We substitute each value and calculate the force:
[tex]\begin{gathered} F=9\times10^9\cdot\frac{(13.13\times10^{-9})(26.7\times10^{-9})}{(0.0272)^2} \\ \\ F=0.004259\text{ N} \end{gathered}[/tex]The magnitude of the electric force on object A is 0.004259 N
A basketball player hit the basket eighteentimes during the game. He scored a totalof 30 points, two points for each field goaland one point for each free throw. If hethrew 6 more field goals than free throws,how many field goals did he make? Howmany free throws did he make?
Answer:
fxn g
Explanation:
ggnhbdfhbz
Bluetooth and Wi-Fi use the same frequency band. Give one reason why Bluetooth signal does not interfere with Wi-Fi signal.
There is interference between both signals especially if they are operated in same time and space. The technology that bluetooth uses helps reduce interference from wifi signals. This technoogy is called the frequency hopping spread spectrum which changes channel frequency. This only reduces interference.
Question 24 (1 point)—A beam of white light hits a prism and is dispersed into the its spectral componentsred, orange, yellow, green, blue, indigo, and violet. If a second prism is placed so thatonly the red band of light can pass through, what will happen?No light will pass through the second prism because red light does not haveenough energyThe prism will disperse the red light into many shades of redThe prism will polarize the red lightThe prism will refract the red light but it will still be the same color of red
prism can disperse a compound light
red is a single coloured light
when it will pass through the prism it will get refracted again but the output will be the red one
so the answer is
The prism will refract the red light but it will still be the same color of red
Jack is selling lemonade at a stand outside his house. He is charging $1.50 per cup. He needs to make at least enough money to cover the cost of his supplies, which were $15.25.What is the minimum number of cups of lemonade Jack needs to sell?
Given,
Charge of one cup of lemonade, c=$1.50
The amount Jack needs, T=-$15.25
The number of cups Jack needs to sell in order to have the given amount can be calculated as,
[tex]N=\frac{T}{c}[/tex]On substituting the known values,
[tex]\begin{gathered} N=\frac{15.25}{1.5} \\ =10.17 \end{gathered}[/tex]The number of cups of lemonade that can be sold should be a whole number. If Jack sells 10 cups, which is lesser than the answer, he will have less amount than he wanted.
Thus he has to sell 11 cups.
Therefore the minimum number of cups Jack needs to sell is 11.
Answer: I have all answers in screen shots
Hope this helps
An element with an atomic number of 90 and an atomic mass of 230 would have what atomic number after Alpha decay?88909186
Given data:
The atomic number of
66) The fundamental wavelength for standing sound waves in an empty pop bottle is 0.88 m. (a) What is the length of the bottle? (b) What is the frequency of the third harmonic of this bottle?
Explanation
Step 1
Let
[tex]\begin{gathered} \text{fundamental wavelength=0.88 ,} \\ \lambda=2L \\ \text{hence, replace} \\ 0.88m=2\text{ L} \\ \text{divide both sides by 2} \\ \frac{0.88m}{2}=\frac{2\text{ L}}{2} \\ 0.44=L \end{gathered}[/tex]hence, the length of the bottle is 0.44 meters
Step 2
What is the frequency of the third harmonic of this bottle?
For the third harmonic, the length of the string is equivalent to three-halves of a wavelength,so
[tex]\begin{gathered} third\text{ armonci= }\frac{\text{3}}{2}\cdot0.88\text{ m} \\ third\text{ armonci= }1.32\text{ m} \\ \end{gathered}[/tex]so
[tex]\begin{gathered} 1.32=2L \\ \end{gathered}[/tex]replace in the quation and solve for
Let speed of sound = 343 m/s
[tex]\begin{gathered} f=\frac{v}{\lambda} \\ \text{replacing} \\ f=\frac{343}{0.88} \\ f=389.77\text{ Hz} \end{gathered}[/tex]I hope this helps
you
A sports car with a mass of 2000 kg travels at a velocity of 20 m/s. What would be true of its momentum if the same sports car was traveling at a velocity of 40 m/s?
The true statement about the sports car is that the momentum of the sports car would be increased.
What is momentum?The term momentum is defined as the product of the mass and the velocity. The momentum of an object is a vector quantity. What it means is that both the magnitude and the direction of the momentum is important.
Now we are told that the sports car has a mass of 2000 kg travels at a velocity of 20 m/s. The momentum at this point is 40000 Kgm/s. If the velocity of the sports car was increased to 40 m/s, the momentum would increase to 80,000 Kgm/s.
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A girl weighs 58kg and runs up a staircase 52m high in 23seconds. Find her minimum power output
Given,
The mass of the girl, m=58 kg
The height of the staircase, h=52 m
The time she takes to run up the staircase, t=23 s
The work the girl does to climb up the staircase is equal to the change in the potential energy of the girl.
Thus the work done is given by,
[tex]W=\text{mgh}[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} W=58\times9.8\times52 \\ =29.56\times10^3\text{ J} \end{gathered}[/tex]The power output is equal to the rate at which the work is done
Thus the power output is given by,
[tex]P=\frac{W}{t}[/tex]On substituting the known values,
[tex]\begin{gathered} P=\frac{29.56\times10^3}{23} \\ =1285.22\text{ W} \end{gathered}[/tex]Thus the maximum power output is 1285.22 W
A 49 lb piece of steel at 670 F is dropped into 95 lbs of water at 50 F. What is the final temperature of the mixture.
ANSWER
[tex]83.82F[/tex]EXPLANATION
Parameters given:
Mass of steel, M = 49 lb = 22.23 kg
Initial temperature of steel, ts = 670 F = 627.59 K
Mass of water, m = 95 lb = 43.09 kg
Initial temperature of water, tw = 50 F = 283.15 K
We want to find the final temperature of the mixture of the piece of steel and water.
The amount of heat released in the system must be equal to the total amount of heat absorbed by the system:
[tex]Q_R=Q_A[/tex]The piece of steel releases heat while the water absorbs the heat. This implies that the equation above becomes:
[tex]-MC(T_s-t_s)=mc(T_w-t_w)_{}[/tex]where C = specific heat capacity of steel = 468 J/kgK
c = specific heat capacity of water = 4184 J/kgK
Ts = final temperature of steel
Tw = final temperature of water
Since the final temperatures of both the steel and water are equal, it implies that:
[tex]T_s=T_w=T[/tex]Substituting the given values into the equation above, we have that:
[tex]-22.23\cdot468\cdot(T-627.59)=43.09\cdot4184\cdot(T-283.15)[/tex]Simplify the equation above:
[tex]\begin{gathered} -10403.64(T-627.59)=180288.56(T-283.15) \\ \Rightarrow-10403.64T+6529220.428=180288.56T-51048705.76 \\ \Rightarrow180288.56T+10403.64T=6529220.428+51048705.76 \\ \Rightarrow190692.2T=57577926.19 \\ T=\frac{57577926.19}{190692.2} \\ T=301.94K \end{gathered}[/tex]Let us now convert to Fahrenheit:
[tex]T=\frac{(301.94-273.15)\cdot9}{5}+32=83.82F[/tex]That is the final temperature of the mixture.
Is science physics? If so I have a question on it
Inner Planets:
A planet whose orbit lies within the asteroid belt is called inner planets. Mercury, Venus, Earth and Mars are called th inner planets.
Outer Planets:
A planet whose orbit lies outside the asteroid belt is called outer planets. Jupiter, Saturn, Uranus and Neptune are called Outer Planets.
If you have a concave mirror whose focal length is 100.0 cm, and you want an image that is upright and 10.0 times as tall as the object, where should you place the object?
given
[tex]\begin{gathered} focal\text{ length=100 cm} \\ let\text{ object distance is u form the pole of mirror } \\ so\text{ u=-u} \\ and\text{ v=?} \\ here\text{ v is the distance of image from the pole of mirror } \\ let\text{ height of the object h} \\ so\text{ h=+h} \\ according\text{ to the question image of the object is } \\ \text{ h}^{^{\prime}}=+10h \\ \end{gathered}[/tex]since the image is in the upright direction so the image is virtual.
and in the case of the mirror, an image can be virtual if and only if
the image is placed between the pole and the focus. so the object
must be placed between the pole and the focus.
[tex]\begin{gathered} the\text{ formula for magnification is given by;} \\ m=-\frac{v}{u}=\text{ }\frac{h^{^{\prime}}}{h} \\ and\text{ the mirror formula is given by the following} \\ \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \end{gathered}[/tex][tex]\begin{gathered} -\frac{v}{u}=\frac{h^{^{\prime}}}{h} \\ by\text{ putting all the values} \\ -\frac{v}{u}=\frac{10h}{h} \\ v=-10u \end{gathered}[/tex][tex]undefined[/tex]It is hard to see the motion of stars because of their extreme distance from us.
It is hard to see the motion of stars because of their extreme distance from us is a True statement.
The locations of the stars in the sky remain largely constant with respect to one another. Taurus is constantly close to Orion. Draco is continually around Ursa Minor. The stars are a continuous backdrop painting when we observe the sky; they don't move much in relation to one another.
While we orbit the sun throughout the year, there are some minute movements. Stars farther away move slightly more left and right than stars closer to us. Even if it has a minor effect, it makes sense. You'll notice that distant items move less in relation to you than nearer ones if you stare out into the distance while you go along the street.
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The Round Up carnival ride below has a radius of 3.90 meters and rotates 0.527 times per second. As shown, riders can be held up by only friction. What coefficient of friction is needed to keep the riders from sliding down? Include units in your answer. Answer must be in 3 significant digits.
Given data:
* The radius of the carnival ride is r = 3.9 m.
* The linear frequency of oscillation is f = 0.527 Hz.
Solution:
The angular frequency (angular speed) of the oscillation is,
[tex]\begin{gathered} \omega=2\pi f \\ \omega=2\pi\times0.527 \\ \omega=3.31\text{ rad/s} \end{gathered}[/tex]The frictional force acting on the ride is,
[tex]F=\mu mg[/tex]The force acting on the ride in terms of the angular speed is,
[tex]F=m\omega^2r[/tex]The frictional force acting on the ride is equal to the force acting on the ride for circular motion,
[tex]\begin{gathered} \mu mg=m\omega^2r \\ \mu g=\omega^2r \\ \mu=\frac{\omega^2r}{g} \end{gathered}[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} \mu=\frac{(3.31)^2\times3.9}{9.81} \\ \mu=4.36 \end{gathered}[/tex]Thus, the coefficient of friction is 4.36.
The barrel of a rifle has a length of 0.983m. A bullet leaves the
muzzle of a rifle with a speed of 602m/s. What is the
acceleration of the bullet while in the barrel? A bullet in a rifle
barrel does not have constant acceleration, but constant
acceleration is to be assumed for this problem.
Answer in units of m/s^2
According to the solving, the acceleration of the bullet while in the barrel is 184335.7 m/s^2.
What is acceleration?Any action where the velocity changes are said to it as acceleration. There are only two ways to accelerate: altering your speed or altering your direction, or altering both. This happens because velocity is both a speed and a direction.
According to the given data,
The length of the barrel of the rifle is 0.983 m.
The speed of the bullet is 602m/s.
We know,
v^2 = u^2+ 2aS
Initial velocity, u=0
v=602m/s.
S= 0.983 m.
So,
v^2=u^2+2aS
(602)^2= 0+ 2*a*0.983
1.966a=362404
a=362404/1.966
a= 184335.7 m/s^2.
According to the solving, the acceleration of the bullet while in the barrel is 184335.7m/s^2.
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The speed of sound in air is 340 m/s. What is the wavelength of a soundwave that has a frequency of 903 Hz?
ANSWER:
0.38 meters.
STEP-BY-STEP EXPLANATION:
Given:
The speed of sound in air (v) = 340 m/s
Frequency (f) = 903 Hz
We calculate the wavelength using the following formula:
[tex]\begin{gathered} \lambda=\frac{v}{f} \\ \\ \text{ We replacing:} \\ \\ \lambda=\frac{340}{903} \\ \\ \lambda=0.37652\cong0.38\text{ m} \end{gathered}[/tex]The wavelength is equal to 0.38 meters.
A block of aluminum weighing 140 g is cooled from 98.4°C to 62.2°C with the release of 1080 joules of heat. From this data, calculate the specific heat of aluminum.Group of answer choices2.11 J/g °C4.29 J/g °C0.213 J/g °C
213 J/g °C is the specific heat of aluminum when a block of aluminum weighing 140 g is cooled from 98.4°C to 62.2°C with the release of 1080 joules of heat.
C=Specific heat
Q=heat energy (J)0.14 K
M =Mass (Kg)
∆T= change in temperature (K)
M= 140 g = 0.14Kg
Q - 1080 Joules.
∆T=98.4 – 62.2 = 36.2
Q=MC∆T
C =Q/M∆T
C= 1080÷0.14×36.2
C=213 J/g °C
The term "specific heat" refers to the quantity that quantifies how a substance's additional heat energy affects its temperature. The unit is J/g °C. It is generally used in Thermodynamics.
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