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.
Answers
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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Related Questions
who has the best answer for this
what percentage more water is used to provide us with electricity vs for irrigation
Answers
Answer:
electricity because it has more percentage nd energy
Explanation:
mark me brainliest pl
A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark. Assume each tick mark represents 1 cm. Calculate the total displacement if a toy car starts at 0, moves 5 cm to the left, then 8 cm to the right, and then 3 cm to the left. The car moves cm, so there is no displacement.
Answers
The total displacement of the toy car at the given positions is 0.
The given parameters;
First displacement of the car, = 5 cm leftSecond displacement of the car, = 8 cm rightThird displacement of the car, = 3 cm to the leftThe total displacement of the car is calculated as follows;
Let the left direction be "negative direction"Let the right direction be "positive direction"[tex]\Delta x = - \ 5\ cm \ + \ 8 \ cm \ - \ 3 \ cm \\\\\Delta x = 0[/tex]
Thus, the total displacement of the toy car at the given positions is 0.
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Answer:0
Explanation:
A 3-column table with 1 row. The first column titled distance travelled (meters) has entry 6. 1. The second column labeled lower track elapsed time (seconds) has entry 4. 92. The third column labeled higher track elapsed time (seconds) has entry 3. 36. Based on the time measurements in the table, what can be said about the speed of the car on the lower track as compared to the higher track? How can the reasoning for the above answer be best explained? On the higher track, the elapsed time is. Calculate speeds for each track. How much faster was the car on the higher track than the lower track?.
Answers
Answer:
B,A,A
Explanation:
Answer:
Other guy is correct b,a,a
Explanation:
What do protoplanets eventually grow into?
Answers
Answer:
a planet
Explanation:
If planetesimals, objects that combined to form protoplanets, were to collide head on in the early solar nebula, they would shatter to the point where they would be unable to grow into a larger size, a protoplanet. A protoplanet is a massive object that will eventually become a planet.
Galois drove 60.0 kilometers due west in 5.00 hours and then drove 43.0 kilometers due north in 3.00 hours.
(a) How far did he travel?
(b) What was his average speed?
(c) What was his displacement?
(d) What was his average velocity?
Answers
Answer:
Explanation:
See the attachment for the details. A right triangle is formed to find the hypotenuse of the two legs consisting of the actual driving distances and times. The hypotenuse gives the vector information for the displacement at the end of 8 hours of driving.
The individual driving times and distances are summed to provide:
(a) How far did he travel?
103 km
(b) What was his average speed?
12.88 km/h
(c) What was his displacement?
73.82 km
(d) What was his average velocity?
9.228 km/h
The magnitude, M, of an earthquake is represented by the equation M=23logEE0 where E is the amount of energy released by the earthquake in joules and E0=104. 4 is the assigned minimal measure released by an earthquake. In scientific notation rounded to the nearest tenth, what is the amount of energy released by an earthquake with a magnitude of 5. 5?.
Answers
An earthquake with a magnitude of 5.5 releases 4.5 × 10¹² J of energy.
What is the magnitude of an earthquake?The magnitude of an earthquake is a number that characterizes the relative size of an earthquake and is based on the measurement of the maximum motion recorded by a seismograph.
If the magnitude of the earthquake is M = 5.5, we can calculate the amount of energy released (E) using the following expression.
[tex]M = \frac{2}{3} log(\frac{E}{E_0} )\\\\E = E_0 \times 10^{\frac{3}{2}M } = 10^{4.4} \times 10^{\frac{3}{2}(5.5) } = 4.5 \times 10^{12} J[/tex]
where,
E₀ is the assigned minimal measure released by an earthquake.An earthquake with a magnitude of 5.5 releases 4.5 × 10¹² J of energy.
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scientific notation. jst 11a
Answers
Answer:
2 * 10^4
Explanation:
(3 * 10^4)(4 * 10^4) / (6 * 10^4)
= (12 * 10^8)/(6 * 10^4)
= 2 * 10^4
The International Space Station (ISS) orbits Earth at an altitude of 4.08 × 105 m above the surface of the planet. At what velocity must the ISS be moving in order to stay in its orbit?
A) 7.91 × 10^3 m/s
B) 3.12 × 10^4 m/s
C) 7.66 × 10^3 m/s
D) 8.17 × 10^3 m/s
Answers
This question involves the concepts of orbital velocity and orbital radius.
The orbital velocity of ISS must be "7660.25 m/s".
The orbital velocity of the ISS can be given by the following formula:
[tex]v=\sqrt{\frac{GM}{R}}[/tex]
where,
v = orbital velocity = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.97 x 10²⁴ kg
R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m
R = 67.86 x 10⁵ m
Therefore,
[tex]v=\sqrt{\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{67.86\ x\ 10^5\ m}}[/tex]
v = 7660.25 m/s
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Do I look good yes or no
Answers
Answer:
0/10
Explanation:
_--------------------
Answer:
No. -infinity/10
Explanation:
Ew. Just why, like no.
An object, initially at rest, moves 475 m in 19 s. What is its acceleration? *
Answers
[tex]\\ \sf\Rrightarrow s=ut+\dfrac{1}{2}at^2[/tex]
[tex]\\ \sf\Rrightarrow s=0(19)+\dfrac{1}{2}a(19)^2[/tex]
[tex]\\ \sf\Rrightarrow 475=\dfrac{361a}{2}[/tex]
[tex]\\ \sf\Rrightarrow 950=361a[/tex]
[tex]\\ \sf\Rrightarrow a\approx 2.3m/s^2[/tex]
A neutral object has:
a. Zero electrons.
b. Zero protons.
c. Only neutrons.
d. An attraction to negative objects.
e. An attraction to neutral objects.
Answers
Answer:
c. Only neutrons.
hope it helps :)
7. If the speed of light in medium-1 and medium-2 are 2.5x 10 m/s and 2x 10 m/s respectively then the refractive index of medium-1 with respect to medium-2 is (a) 3/2.5 (b) 2/2.5 (c) 2.5/3
Answers
Answer:
Answer to the question is 4 / 5
To know the answer with proof see the above attachment"true wisdom is knowing what you don't know"
Reader's Response/explanation
(Y'all I need help)
Answers
Answer:
“The only true wisdom is in knowing you know nothing.” ― Socrates. Socrates had it right. Remain humble and appreciate the fact that you don't know everything. ... Thankfully, one's search for knowledge has no end as there is always something new to learn, giving one's life meaning and purpose.
Explanation:
Answer:
knowing wisdom cool
Explanation:
lol
The distance between adjacent nodes in a standing wave pattern is 25.0 cm. What is the
wavelength? If the frequency is 200. Hz, what is the speed of the wave?
Answers
Answer:
Answer:
Speed of the wave in the string will be 3.2 m/sec
Explanation:
We have given frequency in the string fixed at both ends is 80 Hz
Distance between adjacent antipodes is 20 cm
We know that distance between two adjacent anti nodes is equal to half of the wavelength
So \frac{\lambda }{2}=20cm
2
λ
=20cm
\lambda =40cmλ=40cm
We have to find the speed of the wave in the string
Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec
So speed of the wave in the string will be 3.2 m/sec
The practice of science can answer only scientific questions. And scientific questions guide the design of investigations. What must be true of the possible answers to a scientific question? A. They are popular with a majority of scientists. B. They agree with all prior experiments. O C. They can be supported by evidence. O D. They lead to increased funding of scientific research.
Answers
Answer:
They must be supported by evidence.
Explanation:
Every scientific theory or scientifc claim must have scientific evidence.
Two 3.0g bullets are fired with speeds of 40.0 m/s
and 80.0 m/s, respectively. What are their kinetic
energies? Which bullet has more kinetic energy? What is the ratio of their kinetic energies?
Answers
[tex] = \sf \frac{1}{2} m {v}^{2} [/tex]
Kinetic energy of the 1st bullet [tex] \sf = \frac{1}{2} \times m1 \times {(v1)}^{2} \\ \sf= \frac{1}{2} \times 0.003 \times {(40)}^{2}J \\ \sf = \frac{1}{2} \times 0.003 \times 1600J \\ \sf = 2.4J[/tex]Kinetic energy of the 2nd bullet [tex] \sf = \frac{1}{2} \times m2 \times {(v2)}^{2} \\ \sf= \frac{1}{2} \times 0.003 \times {(80)}^{2}J \\ \sf = \frac{1}{2} \times 0.003 \times 6400J \\ \sf = 9.6J[/tex] So, the 2nd bullet which has greater velocity has more kinetic energy.Therefore, the ratio of their kinetic energies[tex] \sf = \frac{2.4J}{9.6J} \\ \sf= \frac{24}{96} = \frac{1}{4} \\ \sf = 1 : 4[/tex]
Answer:
The kinetic energies of the bullets are 2.4 J and 9.6 J.
The bullet having greater velocity has more kinetic energy.
The ratio of their kinetic energies is 1 : 4.
Hope you could get an idea from here
Doubt clarification - use comment section.
The kinetic energy of the two bullets are 2.4 J and 9.6 J respectively.
The ratio of the kinetic energy of the bullets is 1:4.
What is kinetic energy?The kinetic energy of an object is the energy possessed by the object due to its motion.
The kinetic energy of the two bullets is calculated as follows;
[tex]K.E_1 = \frac{1}{2} \times 0.003 \times 40^2 = 2.4 \ J\\\\K.E_2 = \frac{1}{2} \times 0.003 \times 80^2 = 9.6 \ J[/tex]
Ratio of the kinetic energy of the bullets is calculated as follows;
[tex]K.E_1 : K.E_2 = 2.4: 9.6 \ = \ 1: 4[/tex]
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If the 0.10 kg baseball on the right has a
kinetic energy of 5 J, what is its velocity?
Answers
Answer:
10 m/sExplanation:
The kinetic energy of an object can be found by using the formula
[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]
m is the mass
v is the velocity
Since we're finding the velocity we make v the subject
That's
[tex]v = \sqrt{ \frac{2k}{m} } \\ [/tex]
From the question
k = 5 J
m = 0.1 kg
We have
[tex]v = \sqrt{ \frac{2 \times 5}{0.1} } = \sqrt{ \frac{10}{0.1} } = \sqrt{100} \\ = 10[/tex]
We have the final answer as
10 m/sHope this helps you
A system consisting of two particles is known to have zero total momentum. Does it follow that the kinetic energy of the system is zero as well?
Answers
Answer:
all of the particles are at rest (v = 0)
Explanation:
Therefore, since nothing is moving, the total momentum of the system must also be zero.
As a system consisting of two particles is known to have zero total momentum, the kinetic energy of the system will be zero as well.
What is momentum?Momentum is the product of a particle's mass and velocity. Momentum is a vector quantity, which means it has a magnitude as well as a direction.
According to Isaac Newton's second law of motion, the time rate of change of momentum equals the force acting on the particle.
Momentum describes the relationship between an object's mass, velocity, and direction.
Any change in momentum generates force. As a result, a change in momentum is used to calculate the force acting on the object.
Because the total kinetic energy is zero, all of the particles are at rest (v = 0). As a result, because nothing is moving, the system's total momentum must also be zero.
Thus, the statement is true.
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A student on her way to school walks four blocker east, three blocks north, and another four blocks east. Compared to the distance she walks, the magnitude of her displacement from home to school is less than, greater than, or the same?
Answers
Answer:
The magnitude of the student's displacement is less than the distance she walked.
Walking: 11 Blocks
Displacement: 8.54 Blocks
Explanation:
See the attached diagram. The unit of length is blocks. We can add the actual blocks walked as shown. She walked a total of 11 blocks.
Her displacement is the distance measured directly from where she started (line A). Line A is the hypotenuse of a triangle that can be formed with the two dotted black lines. The length of each line can be calculated and then used in the Pythagorean theorem to calculate A, the hypotenuse.
That result is 8.54 Blocks, a shorter distance, once she earns her wings.
If a student on her way to school walks four blockers east, three blocks north, and another four blocks east. Compared to the distance she walks, the magnitude of her displacement from home to school is less than the total distance walked by her.
What is displacement?An object's position changes if it moves in relation to a reference frame, such as when a passenger moves to the back of an airplane or a professor moves to the right in relation to a whiteboard.
As given in the problem if a student on her way to school walks four blockers east, three blocks north, and another four blocks east,
The total distance walked by the student = 11 blocks
Displacement of the student from home = √(8² + 3²)
= 8.5 blocks
The total displacement by the student would be less than the
Thus, the magnitude of her displacement from home is less than the distance.
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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?
Answers
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
What is the work done in lifting 60 kg of blocks to a height of 20m
Answers
Answer:
The answer is 1200
Explanation:
A 300 g ball swings in a vertical circle at the end of a 1.3-m-long string. When the ball is at the bottom of the circle, the tension in the string is 13 N.
What is the speed of the ball at that point?
Answers
Answer:
0.23N for the speed
at the bottom of the circle
Bill pushed 327 kg. bucket of concrete with a force of 10N. What was the
acceleration?
Answers
Answer:
F(10)=mass(327)x acceleration(m/s)
Explanation:
Give proof that mass is constant and weight keeps changing.
Answers
The mass of an object stays the same wherever it is, but its weight can change. This happens if the object goes where the gravitational field strength is different from the gravitational field strength on Earth, such as into space or another planet.
the image in figure 16.36 combines observations made with visible light and radio telescopes. which color in the image represents the radio emission?
Answers
Answer: pinkish- white
Purple
Explanation:
Got a question ~ can someone help me out ?
I want Explanation ~ [ cuz there's something I'm not clear about in this question ]
Thanks for Answering ~
Answers
Answer:
Hey Tyere!
Refer to the attachments....
I hope it is helpful to you...Cheers!____________a 25 kg crate is pushed horizontally with a force of 70N. if the coefficient of friction is .20, calculate the acceleration of the crate.
Answers
From Newton's second law of motion, the acceleration of the crate is 0.84m/[tex]s^{2}[/tex]
Given that a 25 kg crate is pushed horizontally with a force of 70N and the coefficient of friction is 0.20
To calculate the acceleration of the crate, We will consider the net force acting on the crate. The net force will be sum of the applied force F and the frictional force [tex]F_{r}[/tex].
Let us first calculate the frictional force [tex]F_{r}[/tex]. That is,
[tex]F_{r}[/tex] =- μN
[tex]F_{r}[/tex] = μmg
[tex]F_{r}[/tex] = 0.2 x 25 x 9.8
[tex]F_{r}[/tex] = 49N
Using Newton's second law,
F = ma
F - [tex]F_{r}[/tex] = ma
70 - 49 = 25 a
21 = 25a
a = 21 / 25
a = 0.84 m/[tex]s^{2}[/tex]
Therefore, the acceleration of the crate is 0.84m/[tex]s^{2}[/tex]
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This property of waves is the only property where the relationship between energy and this property are indirect or inverse
Answers
Answer: I don't understand
Explanation:
study and pay attention
The diagram shows a nephron.
A nephron. W points to branch of renal artery. X points to branch of renal vein. Y points to tubular point of nephron. Z points to collecting duct.
Where is the blood first filtered?
Answers
Answer:
w
Explanation:
renal artery...the blood flow into the kidney via this blood vassles
and the filtration take place is called ultrafiltration
what is momentum?
what is period of an oscillating body?
what is friction?
Answers
Answer:
Explanation:
Momentum is the quantity of motion of a moving body, measured as a product of its mass and velocity.
The period of an oscillating body is the smallest interval of time in which a system undergoing oscillation returns to the state it was in at a time arbitrarily chosen as the beginning of the oscillation.
Friction is a force between two surfaces that are sliding, or trying to slide, across each other. Friction always slows a moving object down. The amount of friction depends on the materials from which the two surfaces are made. The rougher the surface, the more friction is produced.
