5

1) The problem runs along the vertical y axis. g = 9.81. Astone is dropped from the top of a building of height h. At aheight 10.21 m is the lower edge of a window ...

Question

1) The problem runs along the vertical y axis. g = 9.81. Astone is dropped from the top of a building of height h. At aheight 10.21 m is the lower edge of a window of height 2.021m . It takes the stone to travel the window in a time 0.04521s. Calculate the height h.2) The problem unfolds on the horizontal 'x' axis. A car Agoes at constant speed 51.8073 km/h and at t = 0 it is at18.0844 km. Car B is at rest at t = 0 at 10.7029 km. Calculate theminimum constant acceleration aB that B mus

1) The problem runs along the vertical y axis. g = 9.81. A stone is dropped from the top of a building of height h. At a height 10.21 m is the lower edge of a window of height 2.021 m . It takes the stone to travel the window in a time 0.04521 s. Calculate the height h. 2) The problem unfolds on the horizontal 'x' axis. A car A goes at constant speed 51.8073 km/h and at t = 0 it is at 18.0844 km. Car B is at rest at t = 0 at 10.7029 km. Calculate the minimum constant acceleration aB that B must have to reach A before it reaches the position 28.8404 km. 3) g = 9.81. The problem unfolds on a plane. The theme is second law of Newton. A block of mass a= 0.122 kg hangs on a rope "c12" that passes through a pulley fixed to the left corner of a horizontal table without friction. Said rope turns the pulley to the right and joins a mass b= 0.526 kg from which comes another rope "c23" to the right that turns into a pulley and goes down joining a mass c= 0.285 that hangs from it. The pulleys have no mass. Find the acceleration a2 of mass b, with the convention of positive if it is to the right, and the tensions of chord c12 and tension of chord c23. 4) Angles in degrees and g = 9.81. The problem unfolds on a plane. A block of mass a= 0.122 kg hangs from a rope "c12" that passes through a pulley fixed to the left corner of a horizontal friction table. This rope turns the pulley to the right and is attached to a mass b= 0.2116952 kg. The pulleys have no mass. The coefficient of kinetic friction between mass b and the plane is 0.226. The mass a falls with acceleration a1. Find the acceleration a1 of mass a, with the convention of positive if downward, and the tension t12 of the chord c12.



Answers

As shown in Fig. $10-3$, a mass $m=400$ g hangs from the rim of a frictionless pulley of radius $r=15 \mathrm{~cm}$. When released from rest, the mass falls $2.0 \mathrm{~m}$ in $6.5 \mathrm{~s}$. Find the moment of inertia of the wheel. The hanging mass linearly accelerates downward due to its weight, and the pulley angularly accelerates clockwise due to the torque produced by the rope. The two motions are linked by the fact that $a_{T}=r \alpha$. Consequently we will need to determine $a_{T}$, and then $\alpha$, and then $F_{T}$, and then $\tau$, and then $I$. Remember that Newton's Second Law is central here (i.e., $\tau=I \alpha$ for the wheel and $F=m a$ for the mass). First we find $a$ using $y=v_{i} t+\frac{1}{2} a t^{2}$, since the mass accelerates down uniformly: $$ 2.0 \mathrm{~m}=0+\frac{1}{2} a(6.5 \mathrm{~s})^{2} $$ which yields $a=0.095 \mathrm{~m} / \mathrm{s}^{2}$, and that equals the tangential acceleration $\left(a_{T}\right)$ of a point on the rim of the pulley, which equals the acceleration $a$ of the rope. Then, from $a_{T}=\alpha r$, $$\alpha=\frac{a_{T}}{r}=\frac{0.095 \mathrm{~m} / \mathrm{s}^{2}}{0.15 \mathrm{~m}}=0.63 \mathrm{rad} / \mathrm{s}^{2}$$ The net force on the mass $m$ is $m g-F_{T}$ and so $F=m a$ becomes $$\begin{aligned} m g &-F_{T}=m a_{T} \\ (0.40 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right) &-F_{T}=(0.40 \mathrm{~kg})\left(0.095 \mathrm{~m} / \mathrm{s}^{2}\right) \end{aligned} $$ from which it follows that $F_{T}=3.88 \mathrm{~N}$. Now $\tau=I \alpha$ for the wheel: $$\left(F_{T}\right)(r)=I \alpha \quad \text { or } \quad(3.88 \mathrm{~N})(0.15 \mathrm{~m})=I\left(0.63 \mathrm{rad} / \mathrm{s}^{2}\right)$$ from which we get $I=0.92 \mathrm{~kg} \cdot \mathrm{m}^{2}$.

Moment and we have a table if detainer that we measure the mess one mess to at the acceleration and we want to get their mind. I quantity that when plotted horizontal axis of graphs should result in a straight line and make the blood. So with remind the the straight line equation and then we just catch what the but should be. So in this case, we can apply the neutrals like waiting for the system and get M one g manifestation is if also m one a empty. Is it going to end to a We use this t here and to solve for the acceleration and this gives us a is you could took em one over m one plus m two g. So if we plot A versus M one over M one plus m two, this should be a straight line. Where this look is the gravity and we can just catch this plant. We have the straight line here and the values Are there a point to you for maximum? Right here is one. And then or the oh, they're access. We have 0.2 and here we have 0.1 So this is more or less what this blood to look like. And the dots will be something like bear around around it. So we have you have to do a linear fit toe, find a plant or just right and all the straight line where a matches most off the points inside. And if you try to copulate, where is this? The slope were found the Slope IHS 9.9 meet us for a second scrap, which is 1.3% higher than the the graphic that these should be the theoretical value for the slope. But this error is inside. The experimental wears off the experiment.

Question number 77 two blobs in that system. The have yet block is on the horizontal surface is where this m a G. And it's very is and be g he had rope is not mass less. So let us take. This portion is off land D and the total length is l. Then we can write the mass off rope off the length is and by L into the This is a moss off the rope off Dylan. So the net force acting on that two blocks and ropes system is MB G Bless the mass off the hanging rope. And by and myself, you are dead by L in Dordain J. This is the net force acting on the system because in the questions, affection is not present. So this force is equal to the net force on the system, which is a Quito net must and may less me less, um, odd into the X elation off the system. So from this relation, we get the X elation off that system a equal to de times MB less Ahmad. They were aided by l in the day they were dead by MB less um a less Ahmad. This is the required isolation off the system. No, in the if the must be falls, then the land off hanging rope, they will increase. And as the length off the hanging robi will increase the mass off the hanging rope also increase and does the net force on the total system increases And hence the acceleration will also in these? No, for that might be. We have to find the length today when the fiction is present. So when the friction is present and the acceleration is zero, that is, the system is just about to move. Then we can write them that force on the system, which is and, uh, be by l less and be indoor g. This that force on the system should be called to the frictional force meal as, um a g so no substituting the values. Marcel rope it zero point 16 in the day, divided by one less 0.4. This is a call to 0.25 in tow. Solving this, we will get the desired. Were you off the length off the hanging rope as zero point 6 to 5 made that now in the last part of the question we have to find but that the system move if the Marsa appropriate 0.4 Getty so substitute that new values in the same expression. As importantly, we can night 0.0 for new months. Off the rope in the bay they were dead by length of the rope. One, let's settle. Born for is it called? Oh, 0.25 And don't go solving this. We will get the Quito 2.5 m. You can see that this 2.5 m is greater than the total length off the road. So this condition, it is not possible because the maximum length off the rope is 1 m and so forth. This makes Mom land off rope. Open meter. The system will not move. Not no.

This question is about an elevator system. So there's an elevator box, it Harrick E g s. Then inside the box, it can fit and number of people. And the people is, uh, we d k g each okay, and then a za counterweight off 150 kg. And then there's also, uh, putting off mass. 280 k g radio, 0.7 m. And so, um, there are six parts in this question and in part A. So I put in this elevator system, Uh, the way that all over the story is by using, uh, energy conservation measure so it disconnect the model before it reaches the the flaw desired. Okay, so part A want to determine the distance d off the car that cost upwards as a function off em using energy conservation. All right. So he so impressed a, um using before that, I am going to call, uh, and being a star mass off the box elevator box, which is 100 kg. Then my ass off the people, which is 80 k g then the mass of the party. Trying educate g. The man's off. The counterweight is, uh, 150 kg. Then we have our previous off. The pulley is 0.7 years. It's okay. I think that's about it. Right. So we'll be using conservation of energy, uh, to suffer party. Okay. On the system. Uh huh. Consisting the box the people and counterweight. Mhm. Corinne I know of. Okay. Okay, so there is no external force acting on the system, So the energy told energy of the system is conserved, so Okay, I plus you, I Is it go to K F five direction. Okay, So the initial kinetic So initially, the council, the the box class, the number of people. Um, right, Yeah. Come to the square, plus, uh, kinetic energy off the pulley, which is Iomega square, half Iomega square, and then the kinetic energy of the counterweight, then see square. Okay, so this is equal to chaos. Is zero class all right? You f eso Suppose this is the eso I have said k u i to be zero in this configuration. So you have off the box is moving up, so it gains. Um, gravitational potential energy. And then the counterweight is moving down with, so it's going thio reduce. Uh, huh? The gravitational potential energy is going to decrease. So you put the minus sign. Okay. So you can start. Mm. Keeping the terms can be So we have half and B plus an M plus N c he square us have. So the police A disk So have more square. Oh, my God. Square is equal to alright. G d and B class and m minus N c. Okay, Okay. So going to get half embi plus and, um plus, uh, and see plus half m b m uh, each way, teacher, uh, he goes to be our times Omega. Okay. Chiti and B plus and m my name's EMC. Okay, you can start putting in the numbers. Um, so and b iss Ah, it's 100 kg. And then and then yes. The mass of the people is 80. Okay, then multiplied by m. M. C is 280 counterweight. It's 950. And then, uh, counterweight is to our 80. He goes toe 98 cool. And B is, uh, 800 class baby and minus 950. Okay, so Oh, yes, The speed. Yeah. Initial speed is given to be tree, so this is going to be tree square. Okay, so we are going to have, um, nine divided by 1926. So yeah. Good thing 18 90 Pass a d n. Is he going to de times? Uh, maybe end minus 150. Okay, so this is equal to, uh, 19 by 19.6 1890 as a d n divide by maybe. And my next 150 king. Yeah, you can simplify. Can calculate what now I would. 19.6 is, but I'm not going to do that. So this is the answer for part A To then parts be, um, evaluate the distance for Be for anniversary too. Okay. When any cause to to Andi is equal to instead. 9/19 0.6 1890 That's 160 If I buy 160 minus 150 Uh huh. How create you get then. 4.1 m. So this is the answer for part B and then in policy happens when End goes to child eighties. Go to nine or 19. 6 1890 plus Uh huh. 960 if I buy 960 minus 150 Okay, so this is equal to 1.62 Yes. Mhm. Okay, then. Parts. The weapons went in with zero case of the years ago. 29 19.6, um, 1890 give up by negative. 115. This is equal to negative. 5.79 m. Okay. Okay, So this is answer for party, then press E. Who? What individual values? Off end. That's the expression In part, A apply so applies for, uh, and is greater or equal to two in seven parts in this question. Okay. Sorry about that. Hey, then process to explain answer toe e. So for an equal zero and one, get the mass off the elevator. It's less than a counterweight. Mhm. So the car exaggerate upward when release our release. Okay. Okay. Then patchy. Um, suppose an infinite number of people could fit on the elevator. What's the value of D? So this is saying that you're approximately let's and to approach infinity. Okay, so the is equal to 91 19.6. Um, so one each. Yeah. 1890 class. ADM you. Bye bye. E and minus 15 are going to right and no in the new Morita and the denominator. Okay, so you look like this. Yeah. Mhm. Alright. So and then as you take into infinity, um, then you get 9/19 0.6 80 derived by 80. Okay? And then you have our what? Yeah. Yeah. So then you get, uh, 0.459 m. Okay. Yeah, that's correct. Yeah. Okay, King. So and that's the That's all for this question and

In this exercise, we have a block that has a mass of 4 kg that is on the top, often inclined plane that has a length off 8 m and makes an angle of 15 degrees with the horizontal. The block is initially address, and then it starts to slide down the plane and the block has accorded trails along behind it. And then, after the block slides down a distance De, uh, 5 m from the top of the in client blame, someone get takes the court that's behind the block and star streaks exert a certain force or the court such that the block stops exactly at the bottom of the incline plain. And our goal is to calculate the force that the person exerts over the cord in two different ways. First, we have to use conservation of any energy methods in order to find it, and then we have to use Newton's laws and the concept of exploration. No orders would find it. So, first of all, in order to find to find it using energy methods, I'm gonna use the concept that the work that's done over a body is equal to the variation in its mechanical energy. Now notice that initially the body is when it to the top off the inclined plane it has and mechanical energy. Initial mechanic letters You off M g h they were Age is you go to eight times sign, um, 15 degrees. This is the initial mechanical energy and the block keeps this energy for 5 m until someone takes the court and starts to exert a certain force over the court. Then the final energy of the block is zero. Because at the bottom of the inclined plane, H is equal to zero and the block has no kinetic energy. So the work is equal to Ah, The final finish of the E f is equal to the variation and enter use which the final energy minus the initial energy. So this is it minus MGI in the work is the force f times the distance L along which the force is exerted and this is equal to Mina's MGH. So the force is minus MGH, divided by and miss 4 kg G is 9 m per second. Squared age is eight. Sign of 15 and l is 3 m. So F is equal to 27 Newt. Then we can solve it using force, method, force, method and basically notice that we have to draw the forced Bagram on the block. So here we have the block. Uh, we have the gravitational force of that that acts upon the block, but we can divide it into two components, one that is tangential to the inclined plane. And it's zip. Which energy? Science data, Of course I'm 15 and the other one that his MGI course I 15. And we also have the force attention force. Or that's just equal to the force of the person exerts over the court. Yes, and we know that if I show you mg science data, mine is the force is equal to am times the acceleration off the block. So the force will be in times G science data. Mine is the exploration. So let's find what the exploration is. Ifill aeration is equal to actually what I'm gonna do it see is, uh, socialise formulas. We have that the final speed squared minus the initial speed square is equal to two times declaration times the distance traveled. Oh, now the five acceleration is just zero. The initial acceleration the zero is equal to the square root of ship toots g times that age and say this the since the block traveled a distance of 5 m along the plane, then its creation and height is sis. Five time is a sign of 15 which is equal to 1.3 meters. So V zero is equal to the square root of two times uh, 9.8 m per second square times out I age which is 1.3 m. This is equal to 5.5 m per second, so it can go back. Teoh the equation that I was writing in black. So we have that a is equal to minus Z zero squared, divided by two out. The zero is just 5.5 m your second and we have two divided by two times 3 m. So a is equal to minus four points 25 m per second squared. This means that F is equal to the Mass M, which is 4 kg times G, which is not going eight meters per second squared times that ah sign or 15 plus 4.25 meters per second squared, So F is equal to 27 unions, and this is the answer to a question in the both. How using both methods.


Similar Solved Questions

5 answers
Point) Give vector parametric equation for the line that passes through the point (5.4, ~3) , parallel to the line parametrized by (4 4+ 20,-4 - 40) :L(t)
point) Give vector parametric equation for the line that passes through the point (5.4, ~3) , parallel to the line parametrized by (4 4+ 20,-4 - 40) : L(t)...
1 answers
Consider #e System %dlfterental equalens 9 = 43 utre A=Ar dt) = (kco) . Corpte #e fnlmental mtrix Yc+)Kc6) Crpte #e invere mtrx Ycojkc) Use #e inverse malix t Cmpwte tA e ~then uce etA W #rd ~e Sloton b Ivp (nt etAv w #e Sdhn 6 IVP ueve 3 =Av ad %t) =V)
Consider #e System %dlfterental equalens 9 = 43 utre A= Ar dt) = ( kco) . Corpte #e fnlmental mtrix Yc+) Kc6) Crpte #e invere mtrx Ycoj kc) Use #e inverse malix t Cmpwte tA e ~then uce etA W #rd ~e Sloton b Ivp (nt etAv w #e Sdhn 6 IVP ueve 3 =Av ad %t) =V)...
5 answers
Gaseou dinitrogen tetroxidc placed In a flask and allowed decompose nitrogen equillbrlum at 100*€ oride At 100- rezcn the value of Kzq 0.212, the concentration af equillbrium dinitrogen tetroxide at 0.155 mol/L; what concentratian nitrogen dioxide equilibrium?
Gaseou dinitrogen tetroxidc placed In a flask and allowed decompose nitrogen equillbrlum at 100*€ oride At 100- rezcn the value of Kzq 0.212, the concentration af equillbrium dinitrogen tetroxide at 0.155 mol/L; what concentratian nitrogen dioxide equilibrium?...
5 answers
X2) / sec? xp (x
x2) / sec? xp (x...
5 answers
Suppose a student creates solution by dissolving 0.5040 g of a non-dissociating, non-volatile unknown solute in 15.22 g cyclohexane. The freezing point change is -2.8*C (a) Calculate the number of moles of solute in the solution. mol(b) Calculate the molar mass of the solute g/molWhich one of the P-dihalobenzenes could this unknown be (p-difluoro- , P-dichloro, P-dibromo- P-difluorobenzene P-dichlorobenzene P-dibromobenzene p-diiodobenzeneP-diiodobenzene)?
Suppose a student creates solution by dissolving 0.5040 g of a non-dissociating, non-volatile unknown solute in 15.22 g cyclohexane. The freezing point change is -2.8*C (a) Calculate the number of moles of solute in the solution. mol (b) Calculate the molar mass of the solute g/mol Which one of the ...
5 answers
Be sure to answcr aIl parts. of m solution mude by dissolving 9.94 gof KI E What are the mole fraction and the mass percent 0.400 L of water (d = L.0O glmL)?mole fraction KImass % KI
Be sure to answcr aIl parts. of m solution mude by dissolving 9.94 gof KI E What are the mole fraction and the mass percent 0.400 L of water (d = L.0O glmL)? mole fraction KI mass % KI...
5 answers
Him (g(*))answers as comma separated lists:) The equations of the asymptotes (Enter your
Him (g(*)) answers as comma separated lists:) The equations of the asymptotes (Enter your...
5 answers
Sequence ID Species Resistant to DDT AJ131759.1 Myzus persicae AJ131760. Myzus persicae Yes AY763097. Cydia pomonella Yes AY763097.2 Cydia pomonella MF804399.1 Musca domestica isolate Resistont Yes MF804402.1 Musca domestica isolate susceptible Y13592.1 Anopheles gambiae Yes Y13592.2 Anopheles gambiae
Sequence ID Species Resistant to DDT AJ131759.1 Myzus persicae AJ131760. Myzus persicae Yes AY763097. Cydia pomonella Yes AY763097.2 Cydia pomonella MF804399.1 Musca domestica isolate Resistont Yes MF804402.1 Musca domestica isolate susceptible Y13592.1 Anopheles gambiae Yes Y13592.2 Anopheles gambi...
5 answers
Use the graph of pdf to answer the following 4 questions(2) Plx-4)=(3) The total area over interval (0, 10) is(4) (4) what is the value of h, the height ofithe rectangle?'(5) The probability represented byithe shaded area is
Use the graph of pdf to answer the following 4 questions (2) Plx-4)= (3) The total area over interval (0, 10) is (4) (4) what is the value of h, the height ofithe rectangle? '(5) The probability represented byithe shaded area is...
5 answers
Q1. Draw the mechanism of Synthesis of (E) and(Z)-1-(4-bromophenyl)-2-phenylethene. What is the rate determiningstep?
Q1. Draw the mechanism of Synthesis of (E) and (Z)-1-(4-bromophenyl)-2-phenylethene. What is the rate determining step?...
5 answers
The energy of a moving vehicle is increasing. How do you know if the work being done on the vehicle is positive or negative? Please explain your answer.
The energy of a moving vehicle is increasing. How do you know if the work being done on the vehicle is positive or negative? Please explain your answer....
5 answers
QuestionMatch the correct answers:A 21C35 D.14 E356.0H.7
Question Match the correct answers: A 21 C35 D.14 E35 6.0 H.7...
4 answers
(1Opts) Find dufdr by implicie difFerentiation. _' AIv - V' =6
(1Opts) Find dufdr by implicie difFerentiation. _' AIv - V' =6...
5 answers
Gelf & 1o} (Tv JapIO 34? JO UOTAEUTOIdde auajyp PIBMIO} puud 8*3
gelf & 1o} (Tv JapIO 34? JO UOTAEUTOIdde auajyp PIBMIO} puud 8*3...
2 answers
For the matrix A, find (if possible) nonsingular matrix such that P-1AP is diagonal. (If not possible enter IMPOSSIBLE:)A =Verify that P-JAP is diagonal matrix with the eigenvalues on the main diagonal.P-IAP
For the matrix A, find (if possible) nonsingular matrix such that P-1AP is diagonal. (If not possible enter IMPOSSIBLE:) A = Verify that P-JAP is diagonal matrix with the eigenvalues on the main diagonal. P-IAP...
5 answers
Data Table 2Balancingat F 2 250_ Magnitude (N) Angle 0.39 0 6 017y-component x-component 0.39ForceMass (€) 39 1/7 125 g1.25250" F042 Net Sum FitFz+Fa 0.04 Fi + Fz Fz =Resultant ForceF,2 + Fyz 0 = tan (FxIe_
Data Table 2 Balancingat F 2 250_ Magnitude (N) Angle 0.39 0 6 017 y-component x-component 0.39 Force Mass (€) 39 1/7 125 g 1.25 250" F042 Net Sum FitFz+Fa 0.04 Fi + Fz Fz = Resultant Force F,2 + Fyz 0 = tan (FxIe_...
5 answers
V 1 1 3Dyparmcnu
V 1 1 3 Dyparmcnu...
5 answers
Determine the pH of a 0.045 M nitrous acid solution Ka of nitrous acid is 4.0 x 10-4 pH (report value to only two decimal places)
Determine the pH of a 0.045 M nitrous acid solution Ka of nitrous acid is 4.0 x 10-4 pH (report value to only two decimal places)...

-- 0.019375--