Block Definition and 998 Threads

I think that the kinetic energy is equal to heat right. Mv^2 / 2 = ms delta T Delta T is 0,64 But I don’t really know if the temperature decreases or increases at the end

Hi, I am currently in the situation where I'm trying to finish my master thesis. I have 65 to 70 pages written but I still have the editing to do. Simulations are nearing the end so I have most of what I need to finish. Unfortunately, I still need to read papers and get more background...

The Block Universe Theory, based on Einstein's Relativity, says that past, present and future exist at the same "time", the universe is static, eternal, and things doesn't change, movement is an ilusion because you have memories from the past. But if the universe don't evolve does this imply...

A block of mass 0.2 kg which slides without friction on a θ = 30° incline is connected to the top of the incline by a mass-less spring of relaxed length of 23.75 cm and spring constant 80 N/m as shown in the following figure. (a) How far from the top of the incline does the block stop? (b) If...

since there is no external force in the x-direction linear momentum can be conserved. Hence I get the equation $$0 = mv^2 - 9mV^2$$ where ##v## is the velocity of B towards the right and ##V## is the velocity of A toward the left. also the conservation of energy gives $$1/2(9m)V^2 + (1/2)mv^2 =...

This video talks about how air bearings can be made using porous materials like graphite. Is it possible to get a comparable performance using non-porous materials, with an array of pin holes? The video talks about how the carbon seat can be easily worked into conforming with the matching...

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This is my attempt of solving it

I tried to solve it via equilibrium equations and I got 92N for block A

The entire mass of the wedge is ##(M+m)## therefore ##F=(M+m)a##. The forces acting on the small mass are its downward weight ##mg## and the normal force with the contact of the wedge therefore I got that ##N=mg\cos\theta##. Similarly the horizontal component is ##N=ma\sin\theta## therefore...

When drawing a diagram of the forces acting on the block, I have the following forces: $$\sum F_x = a_x = (g \sin\theta) \cos \theta .$$ Now, I can use the following kinematic equation $$x=vt+\frac{a_xt^2}{2}$$, where $$v=0$$ and $$a_x = (g \sin\theta) \cos \theta$$ $$\therefore \frac{2x}{t^2}...

When drawing a diagram of the forces acting on the block, I have the following forces: $$\sum F_x =mg\sin\theta = ma$$ $$g\sin\theta = a$$ however the solution has $$F_x = ax = (g \sin\theta) \cos \theta$$ but I am not sure how they got that? I know the normal force is $$N=mg\cos\theta$$ but the...

the maximum value of friction between the surfaces of the blocks is ##60N## the friction should be self-adjusting until this maximum value. Hence the force of friction in the given scenario should be = 30N but this is physically impossible as if this is the case the top block will be at rest and...

when a force of 10N is applied to the 4kg block the force of friction between the two blocks will also equal 10N as the maximum value of friction between the two surfaces is f = N##\mu## = 20N. if you look at this free body diagram f = F = 10N so the net force acting on the top(4kg) block will...

if the tiny block moves downward by an amount x, the wedge should also move forward by the same amount x as they are connected by the same string whose length has to remain constant, (by differentiating it wrt time we get speed) hence I concluded that v1 = v2, but my book says otherwise what is...

Clearly, in the picture I can see that on the small block on top, tension and gravitational force act. Gravity gets balance by the normal force, so tension is the only force causing acceleration. On the block at the side gravity and tension result in vertical acceleration. However, I do not...

So, for this question first I did a free body diagram in the perpendicular x-y axes, and, I got some equations with the normals, but the answer is independent of normal forces. So, I'm not sure how to eliminate the two normals. Further I find it quite weird that big block-A doesn't come into the...

A gun is fired vertically into a block of wood(unknown mass) at rest directly above it. If the bullet has a mass of 24.0g and a speed of 310 m/s, how high will the block rise into the air after the bullet becomes embedded in it?

We have a wedge whose surface is ##\theta## from the horizontal surface. After a block is placed on its frictionless slant surface, the wedge starts to accelerate due to a force F. What is the normal force acting upon the block? I have been trying to solve it but I got no clue. Could someone...

For question b, the official solution sets up a non-inertial coordinate on the block and writes out the following two equations: $$\begin{cases} \begin{align*} f\cos(\theta)+N\sin(\theta)-mg=0 \qquad \hat\jmath \\ N\cos(\theta)+f\sin(\theta)=ma \qquad\quad\;\;\, \hat\imath \end{align*}...

Diagram attached at the endI personally think there's something wrong with this question, and I'd like if someone can tell me whether it's the question that is wrong or my approach. If I attempt the solution thinking that M should be stationary, the solution is simple. 0 - 1/2 mv^2 = -mgh...

I used to successfully block ads on YouTube with ABP. Now, I'm seeing them. Could I have a virus? Or, does ABP simply not block stuff on Google/Alphabet platforms anymore?

Question 1: I have used v= Aω*cos(ωt+δ) where A= 0.2 m, ω= π/3, t=1 and δ=0. Are the values right in this case? I am confused. Question 2: From question 1 I have got the value of V which is 9 m/s. By using v= ω√(A^2-x^2), I have got the value of x. Now, do I need to add it with 2.5(distance...

Basically, there is a gravitational field g and a friction force acting on the cylinder, which does not slip in the plane. The question is, the acceleration of m'. I solve this question basically by this: " m*g*dh*sina + m'*g*dh = m*v1*dv1 + m'*v2*dv2 + I*w1*dw1 " {1} w*r = v1' + v2' So The...

I thought i could apply a conservation of momentum in this case, Apparently, this is not right, so i don't know what to do now.

This is a rough sketch of the model. It is frictionless. I originally tried simple right-angle trig (sin θ =opp/hyp), but that just ends up as sin θ = sin θ , as well as cos θ = cos θ. I feel like there's also a way to manipulate dynamic equations around to equate something that is capable...

I tried solving this by assuming the acceleration of the truck and block to be the same so the block would stay on the incline. Also, I would assume truck ma = static friction, block ma = mgsintheta... then I solved for a to plug into 1st equation to get 12990 N. Is this correct? I wasn't sure...

Usually questions I have of this nature are down to my limited understanding of the concepts, and this may be no different. Based on my understanding of the Block Universe interpretation of relativity, it appears to me as though there is a conflict between Bell tests and the Block Universe. To...

Just a heads up, this post is quite long. I've tried to be as detailed as possible from the outset because I find it can help avoid the need to clarify things later, or helps when clarifying things later. There is only one question posed at the end, but I think it might be useful to read the...

I don't understand the problem. Does not block A and B make a system, so they should have the same velocity and acceleration at all time? If not, why do they have different accelerations? I don't understand this part of the problem either: "pull applied to block B equals 12.0 N, then block B has...

hello there, I am wondering if my attempt at a solution for this particular problem correct?? I mean I know that the blocks won't move and the total friction force is 38.95N (at least lesser than that, since that is the maximum static friction). I saw someone attempted the solution but it is...

I think that the only force acting on the wall is the normal force caused by Coriolis force, so it can be calculated this way: ##N=m2\dot r \dot \theta## But ##\dot r## is not constant, so how can I calculate it? Then, I can't calculate the acceleration either since I don't have the value of...

Let me draw a free body diagram for the man. I assume that for both cases (a) and (b), the man pulls the block up with uniform velocity. (1) In case (a), realising that the man is in equlibrium (##\Sigma F = 0##) the normal reaction on the man ##N_a = w_M + T## is given by ##N_a = (50+25)\times...

(So this is the system given) The following is my analysis: (i) (ii) Well, my problem is - I got a negative acceleration and its quite impossible to have block B moving to the left. So I am wondering if there are any mistakes I've made.

Change in KE = Change in thermal energy 0.5 * (6)* vblock^2 = 0.4 * 6 * 9.81* 0.1 vblock = 0.885 By Conservation of Momentum, (0.05)(854) = (0.05)*vbu + (6)(0.885)I am not sure whether Change in KE = Change in thermal energy is true coz there should be a change in internal energy of the block...

Summary:: I have done parts a-d but I can't seem to get e correct! A 6.10 kg block is pushed 9.00 m up a smooth 38.0 ∘ inclined plane by a horizontal force of 78.0 N . If the initial speed of the block is 3.20 m/s up the plane. a. Calculate the initial kinetic energy of the block. (found...

stuck on this question

I am using the following code. It's returning the block matrix (Z) raised to negative one (think about inputting 22/7 in a Casio fx-991ES PLUS). import sympy as sp from IPython.display import display X = sp.Matrix([[1, 1, 1], [2, 2, 2], [3, 3, 3]]) i = sp.Matrix([[1], [1], [1]]) Z =...

I know how to solve the problem but have a question related to it. When the block is initially released from the spring the spring, having been pulled back, should give an initial velocity to the block. In that case why is the block's initial velocity zero?

Is it ##(n-1)d^{1->10}ns^2 \ or \ (n-1)d^{1->10}ns^{0->2}## ? My textbook says it is the latter. But I feel like it should be the former. Moreover, period 6 and 7 has the presence of f-orbitals. So adding## (n-2)f^{0->14}## would make it more general. Isn't it? And also what about the f-block...

I calculated the system's change in gravitational potential energy through the equation k=mgh. I used this equation twice, once for the 3.5 kg block with k =(3.5)(9.8)(-0.54) and once for the 1.0kg block, k = (9.8)(1.0). I got -18.522J & 5.292. Is the change just -18.522 - 5.292 J...

Since the problem gave me the kinetic energy and inertia of the iron block, I could plug it into the equation K = 1/2 mv^2 to get the final velocity, I got sqrt(30) for the block's final velocity. From there, I don't really know where to go, I could plug it into the equation above, but ΔE int is...

I tried to solve this question using conservation of momentum. The momentum of the system is mc(vc)=mcvc'+mbvf'. But after that I have no idea I want to use the conservation of kinetic energy but the question doesn't say it's elastic collision, I need to find the velocity of the mb after the...

When the box travels a ## X## distance, the wedge travels ## \frac{X}{2}##. So ##a = 2A## Using the wedge as a non inertial frame: I didn't use (4). Using (2) on (3) and then on (1) I got: ##2mA=mgsin\alpha +mAcos\alpha + \frac{-mgcos\alpha sin\alpha +mAsin^2\alpha +MA}{2cos\beta -...

Summary:: A block of mass m is dropped onto the top of a vertical spring whose force constant is k. If the block is released from a height h above the top of the spring, a) what is the maximum energy of the block? b) What is the maximum compression of the spring? c) At what compression is the...

I first plugged my given values into m1v1+m2v2=(m1+m2)vf. (0.002)(600)+(5)(0)=((0.0020)+(5))vf vf=0.24 m/s Next, I plugged my given values into F=ma. ((0.002)+(5))(9.8) F=49.02 N Next, I plugged my given values into Fdeltat=mdeltav. deltat=mdeltav/F ((0.002)+(5))(0.24)/(49.02)...