Explanation:
The underlying physics give rise to periodic behavior. Usually, some resonance is involved, in which the media damps waves except those at a specific frequency.
__
Only waves at a specific frequency are recognized as being "stationary." If the frequency varies, the behavior is described as "chaotic." So, it's partly a matter of definition.
A dart is thrown at a target that is supported by a wooden backstop. It strikes the backstop with an initial velocity of 350 m/s [E] The dart comes to rest in 0.0050 s.
a. What is the acceleration of the dart?
b. How far does the dart penetrate into the backstop?
Answer:u=350m/sec
v=0m/sec
t=0.005sec
a=v-u/t
=0-350/0.005
= -350/0.005
= -350×1000/5
= -70×1000
= -70000 m/sec^2
Explanation: i think this will help you thankyou
ONLY ANSWER IF YOU KNOW FOR SURE PLEASE :)
answer all three please, it should not take long
Answer:
Question one is b, Question two is b, and question three is b, im pretty positive that what it is
The relationship between Acceleration and Mass is described as:
O a. Equal
O b. Directly Proportional
O c. Inversely Proportional
O d. Opposite
O e. None of these
Explanation:
According to Newton's second law
[tex]\\ \sf{:}\dashrightarrow Force=Mass\times Acceleration [/tex]
[tex]\\ \sf{:}\dashrightarrow Mass=\dfrac{Force}{Acceleration}[/tex]
[tex]\\ \sf{:}\dashrightarrow Mass\propto \dfrac{1}{Acceleration}[/tex]
Option C
A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center, as shown above. The coin doesn’t slip and the time it takes for the coin to make a complete revolution is 1.5 s.
A)The figure below shows the disk and coin as viewed from above. Draw and label vectors on the figure below to show instantaneous acceleration and linear velocity vectors for the pin when it is at the position shown below.
B)Determine the linear speed of the coin
C)The rate of rotation of the disk is gradually increased. The coefficient of stats if friction between the coin and the disk is 0.50. Determine the linear speed of the coin when it just begins to slip.
D)If the experiment in part c were repeated with a second, identical coin glued to the top of the first coin, how would this affect the answer to part c? Explain.
A) Figure attached below
B) The linear speed of the coin = 0.59 m/s
C) Linear speed as coin begins to slip = 0.83 m/s
D) The tangential speed will remain the same as seen in part C
Given data :
mass of coin = 0.0050 kg
Distance of coin from the center of disk = 0.14 m
Time to make a complete revolution = 1.5 s
A) Diagram showing the vectors on the figure is attached below
B) Determine the Linear speed of the coinLinear speed of coin = 2 * π * ( 0.14 ) / 1.5
= 0.59 m/s
C) Determine the linear speed of the coin when it just begins to slipgiven that: friction between coin and disk = 0.50
Friction becomes maximum when coin begins to slip
Maximum frictional force (Fmax) = uV
where V = mg
∴ Fmax = u*mg ---- ( 1 )centripetal force = [tex]\frac{mv^{2} }{r}[/tex] ---- ( 2 )Equating equations ( 1 ) and ( 2 ) to determine the linear speed ( v )
v² = u*r*g
∴ v = √(u*r*g ) = √( 0.5 * 0.14 * 9.8 )
= 0.83 m/s
D) If the experiment is repeated with a second coin glued to the top of the first coin the tangential speed will remain the same
Hence we can conclude that The linear speed of the coin = 0.59 m/s Linear speed as coin begins to slip = 0.83 m/s , The tangential speed will remain the same as seen in part C
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Lower chamber of Congress has how many members?
Answer:435 Representatives
Explanation:The lower chamber of Congress, in which the number of representatives per state is determined by the state's population, with 435 Representatives total. Members of the House of Representatives serve two-year terms, so they are up for reelection every two years.
If astronomers discovered a new planet and found its period of rotation around the Sun to be 105 years, how long would its semi-major axis length be as it orbited the Sun in AU?
From Kepler's third law, its semi-major axis length will be 22.2 AU approximately as it orbited the Sun in AU. The closest option is option C
Given that an astronomers discovered a new planet and found its period of rotation around the Sun to be 105 years.
According to Kepler's third law,
[tex]T^{2} \alpha r^{3}[/tex]
Where
T = Period ( in earth years) = time to complete one orbit
r = Length of the semi major axis in Astronomical unit.
[tex]T^{2}[/tex] = [tex]\frac{4\pi ^{2} }{GM} * r^{3}[/tex]
convert years to seconds
105 x 365 day x 24 hours x 3600 s
T = 3311280000 seconds
Mass of the sun M = 1.989 × 10^30 kg
G = 6.67 x [tex]10^{-11}[/tex]N m^2/kg^2
Substitute all the parameters into the formula
[tex]T^{2}[/tex] = 1.096 x [tex]10^{19}[/tex] = [tex]\frac{4\pi ^{2} }{6.67 * 10^{-11} * 1.989 * 10^{30} } * r^{3}[/tex]
1.096 x [tex]10^{19}[/tex] = 2.976 x [tex]10^{-19}[/tex] [tex]r^{3}[/tex]
[tex]r^{3}[/tex] = 1.096 x [tex]10^{19}[/tex] / 2.976 x [tex]10^{-19}[/tex]
[tex]r^{3}[/tex] = 3.68 x [tex]10^{37}[/tex]
r = [tex]\sqrt[3]{3.68 * 10^{37} }[/tex]
r = 3.33 x [tex]10^{12}[/tex] m
1 AU = 1.5 x [tex]10^{11}[/tex] m
r = 3.33 x [tex]10^{12}[/tex] / 1.5 x [tex]10^{11}[/tex]
r = 22.18 AU
Therefore, its semi-major axis length will be 22.2 AU approximately as it orbited the Sun in AU. The closest option is option C
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Answer:
C. 22.3 AU
Explanation:
Not only is the above an unnecessarily complicated answer, it's not even fully correct, and definitely not what they want you to do.
T^2 = s^3, where T = orbital period and s = semi-major axis length.
Substitute T and you get 105^2 = s^3. Solve for s.
11025 = s^3
3√11025 = s
22.25663649 = s
Therefore, the answer is C. 22.3 AU
How did Thomson's model get its name?
Answer:
thats actually a really good question. sadly i don't know the answer to it. but im sure that there are others who can!
A student lifts a backpack straight up with a force of 53.5 N for a distance of 0.65 m. How much work is done on the backpack?
Happy Christmas Eve!!
Answer:
W = 34.775J
Explanation:
W = F × d
W = 53.5N × 0.65m
W = 34.775J
W = work = unknown
f = force = 53.5N
d = distance = 0.65m
The planets never travel in a straight line as they orbit the Sun. According to Newton's second law of motion, this must mean that _________. the planets are always accelerating a force is acting on the planets the planets have angular momentum the planets will eventually fall into the Sun
Explanation:
The planets never travel in a straight line as they orbit the Sun. According to Newton's second law of motion, this must mean that: Your Answer: The planets have angular momentum.'
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According to Newton's second law of motion, this must mean that force is acting on the planets. The correct option is (b).
As per Newton's second law of motion, an object's acceleration is inversely related to its mass and directly proportional to the net force exerted on it.
The curving courses of the planets in their solar system show that they are constantly changing their direction, as it causes them to accelerate.
The planets are being accelerated in the direction of the Sun as a result of the Sun's gravitational pull on them. The planets continually speed up toward the Sun due to the centripetal force required to maintain them in their orbits, which is caused by the gravitational attraction between the Sun and the planets. As a result, the planets orbit the Sun in elliptical patterns rather than in a straight line.
Hence, According to Newton's second law of motion, this must mean that force is acting on the planets. The correct option is (b).
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The complete question is:
The planets never travel in a straight line as they orbit the Sun. According to Newton's second law of motion, this must mean that _________.
(a) the planets are always accelerating
(b) a force is acting on the planets
(c) the planets have angular momentum
(d) the planets will eventually fall into the Sun
An eagle flying at a constant 120 km/h and has kinetic energy of 2,800 J. What is the mass of the eagle?
Answer:
The mass of the eagle is about 5 Kg
Explanation:
1/2 M= Ke/V^2
120 km/h = 33.3333m/s
1/2 M = 2,800/33.3333^2
1/2 M = 2,800/ 1111.10888889
1/2 M = 2.52000504001
(2) 1/2 M = (2) 2.52000504h001
= M = 5.04001008002
About 5 Kilograms
Describe the water cycle. Be sure to include the following terms in your response: evaporation, condensation, and precipitation.your respone:
Evaporation, or turning liquid to a gas, is first. Then it goes into the atmosphere and codensates, or turns back to a liquid, these water droplets form into clouds and come down as precipitation, rain, sleet, snow etc. Then the water cycle starts again
Initial Velocity is 27.5 m/s. Time is 42 seconds. Final Velocity is 4.5 m/s. Solve for acceleration.
Answer:
-0.5476
Explanation:
a=(Vf-Vi)/t
a=(4.5-27.5)/42
a=-0.547619047
acceleration≈-0.5476
what do i do on this rubixs cube
Answer:
You solve it
Explanation:
Answer:
you just spin it and spin it and spin it and spin it
and spin it
and spin it
and spin it
and spin it
until you get mad then throw it at the wall
Explanation:
ya then you look up a vid that is called rubix cube meme and there should be a pink animal its funny you should watch it!!!!!!!!!
The weight of a person in an elevator at rest = 500 N. Acceleration due to gravity is 9.8 m/s2. When lift accelerated, the tension force is 750 N. What is the acceleration of lift.
Answer:If the elevator accelerated downward then the tension force smallest then 500 N. Otherwise, if the elevator accelerated upward then the tension force larger then 500 N.
The tension force = 750 N because the elevator accelerated upward. Force acts upward has plus sign and force acts downward has minus sign.
T – w = m a
750 – 500 = 50 a
250 = 50 a
a = 250 / 50
a = 5.0 m s–2
Explanation:
Please I need help with this ❤️ ?
Answer:
Compound 'A' C5H12 does not react with phenyl hydrazine. Oxidation of 'A' with K2Cr2o7,/H" gives B' (c5H10o). Compound 'B' reacts with phenyl hydrazine but does not give Tollen's test. The
Put the waves in order from shortest to longest wavelength
Answer:
b, a, c
Explanation:
The middle one has the shortest wavelength, then it's the top one and the last one has the longest wavelength.
Why would poor clusters of galaxies be more likely to have irregular shapes then rich
clusters
These Milky Way companion galaxies are easily visible from dark locations in the Southern Hemisphere. Prime examples of erratic galaxies are the Large and Small Magellanic clouds (left and right, respectively).
What clusters of galaxies likely to have irregular shapes?In comparison to a rich cluster, the poor cluster typically has a slightly more erratic shape. A number of smaller galaxies orbit each major spiral. The Small and Large Magellanic clouds are the two most well-known examples of atypical galaxies.
When two galaxies collide, irregular galaxies frequently result. This unusual Cartwheel Galaxy was created when a tiny galaxy slid through the centre of a massive spiral galaxy.
Therefore, Rich clusters are other clusters that include hundreds to thousands of galaxies. A weak cluster can't cling to its members strongly because of its low bulk.
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NEED HELP ASAP 40 POINTS L!!!! Jaden is interested in learning more about the basics of astrono ny and looking more closely at the sky all by himself. He wantS to get a basic starter telescope to help him with this endeavor, and his teacher SuggestS an organization that might be able to help him with this. What organization is Jaden's teacher likely referring to? The Little Astronomers Celestial Sightseeing Group The Beginning Astronomy Organization Astronomers Without BordersS
please someone help me understand this question
[tex]\\ \sf\longmapsto F=\dfrac{Gm_1m_2}{r^2}[/tex]
m1 and m2 has same units let both be m[tex]\\ \sf\longmapsto F=\dfrac{Gm^2}{r^2}[/tex]
[tex]\\ \sf\longmapsto G=\dfrac{Fr^2}{m^2}[/tex]
SI units:-
F=Newton (N)r=metre(m)mass=m=kg[tex]\\ \sf\longmapsto G=\dfrac{Nm^2}{kg^2}[/tex]
Option C is correct
Which two changes would decrease the electric force between two charged
particles?
- A. Decrease the charge of one of the particles.
B. Increase the charge of both particles.
C. Increase the charge of one of the particles.
D. Decrease the distance between the particles.
E. Increase the distance between the particles.
PLEASE HELP I KEEP FAILING!!!!
Answer:
it is b and e
Explanation:
if u look at the words twice you will notice that b and e are both saying the same meanings just in diff rent words way u need to look close on things like that and u will get passing gradesA strong weightless rope has a mass, m, hanging from the middle of it. The tension force on each rope is 25 N, and the rope droops at an angle of 20.0 degrees. How much mass is hanging from the rope?
By using Lami's theorem, Mass m = 1.75 kg approximately
Given that a strong weightless rope has a mass, m, hanging from the middle of it. If the tension force on each rope is 25 N, and the rope droops at an angle of 20.0 degrees to the horizontal.
By using Lami's theorem, we can get how much mass is hanging from the rope.
Let the angle between the rope = α = 180 - 40
α = 140 degrees
The angle between one of the rope and mass = β = 20 + 90
β = 110 degrees
The angle between the mass and the other rope = γ = 360 - (140 + 110)
γ = 360 - 250
γ = 110 degrees
W/ sinα = T/ sinβ = T/sinγ
W/ sinα = T/ sinβ
Substitute all the necessary parameters
W/sin140 = 25/sin 110
W / 0.643 = 25 / 0.939
W = 17.1 N
Weight W = mg
17.1 = 9.8m
mass m = 17.1/9.8
Mass m = 1.7455 kg
Mass m = 1.75 kg approximately
Therefore, 1.75 kg mass is hanging from the rope.
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A Person is carrying 6kg in one hand and 5kg in another.Calculate the resultant force applied by him
An Olympic runner completes the 200-meter sprint in 23 seconds. What is the runnera''s average speed? (Round your answer to the nearest tenth of a meter per second. ) 0. 9 m/s 1. 2 m/s 8. 7 m/s 10. 1 m/s.
The average speed of the runner is 8.7m/s
Hence, Option C) 8.7m/s is the correct answer.
Given the data in the question;
Distance covered by runner; [tex]d = 200m[/tex]Time taken; [tex]t = 23s[/tex]Average speed; [tex]s =\ ?[/tex]
Speed is simply the rate at which a particle covers a given distance. It is expressed as:
Speed = Distance / Time
[tex]s = \frac{d}{t}[/tex]
We substitute our given distance into the equation;
[tex]s = \frac{200m}{23s} \\\\s = 8.7m/s[/tex]
The average speed of the runner is 8.7m/s
Hence, Option C) 8.7m/s is the correct answer.
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If an input force of 202N is applied to the handles of the wheelbarrow in the sample problem how large is the output force that just lifts the load?
For this case, the relationship between the two forces is given by:
F1 = nF2
Where,
F1: output strengthF2: input forcen: mechanical advantageThen, replacing the values we have:
F1 = (2.2)(202)
Having the calculations we have:
F1 = 444.4N
Answer: The output force that only lifts the load is F1 = 444.4N.
David delivers meals to elderly people once a week. He uses a cart to move the meals. The cart has four smooth wheels. Which type of friction acts between the cartwheels and the sidewalk?
Answer: rolling friction
Explanation: I think it is the answer
Assertion: In domestic electric circuits, the metallic body is connected to the earth wire, which
provides a low-resistance conducting path for the current.
Reason: It ensures that any leakage of current to the metallic body of the appliance keeps its
potential to that of the earth, and the user may not get a severe electric shock. pls fast
Answer:
The answare is it ensures that any leakage of current to the metallic body of the appliance keeps its potential to that of the earth, and the user may not get a severe electric shock.
How charged objects can affect other objects without touching them. ?
Can any1 tell if my answer is right
Answer:
The correct answer would be C. 5.0kg
Explanation:
The mass of an object never changes unless parts of the object are taken away. In other words, although the gravitational force is different on the moon then on the earth the mass of the object would remain the same.
True or false organisms only compete with their own species?
Answer: yes
Explanation:
Animals of different species typically compete with each other only for food, water and shelter. But they often compete with members of their own species for mates and territory as well.
A boxer throws a punch with a force of 1,400 N that lasts 0.02 s. What is the impulse of this punch? (1 point) 28 kg⋅m/s 28 kilograms times meters per second 280 kg⋅m/s 280 kilograms times meters per second 70,000 kg⋅m/s 70,000 kilograms times meters per second 7,000 kg⋅m/s
The impulse of the boxers punch is 28 kgm/s.
The given parameters;
applied force by the boxer, F = 1400 Ntime of force action, t = 0.02 sThe impulse of the boxers punch is calculated as follows;
[tex]J = Ft[/tex]
where;
F is the applied force (N)t is the time of force action (s)The magnitude of the impulse is calculated as follows;
[tex]J = 1400 \times 0.02 \\\\J = 28 \ kg m/s[/tex]
Thus, the impulse of the boxers punch is 28 kgm/s.
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