Physics question: A roof tile falls off the roof of a house. An observer inside the house notices...?
because you are able to't insure for an act of God (to be able to communicate). this is wind - you are able to't insure hostile to the wind. dangle on, feels like that is slightly daft doesn't it - properly it really is. you are able to't insure hostile to the wind. the cost of repairing the roof will be lined below typhoon damage (depending upon the wind speeds) in spite of the indisputable fact that the ensuing damage to the autos in simple terms isn't. Wind is wind and not in any respect an insured peril. this is maximum unlucky the wear and tear to the autos yet you are able to't anticipate to get conceal for this variety of difficulty. the really way you may want to pursue it truly is that if the unique roof craftsman who finished the artwork to the roof replaced into negligent in his artwork, and it would properly be proved that this contributed to the loosening of the tiles - then you certainly'd have an street to pursue. yet except that, you're maximum appropriate selection is to really pay your persons excesses - is it properly worth saving £500 to threat falling out along with your persons?
Physics.. mass slide down roof, final velocity?
Two ways to go, you can calculate the acceleration (using net forces) or just use conservation of energy equations: 1) Calculate acceleration: Fg = mgsin(theta) force of friction = mu * Normal force Fn = mgcos(theta) so net force: Fnet = mg * (sin(theta) - mu * cos(theta)) ("mu" is the greek letter mu denoting coefficient of kinetic friction NOT m * u) then a = Fnet / m = g * (sin(theta) - mu * cos(theta)) then you can use one of the kinematic equations to find final velocity: a * d = (1/2) * (vf^2 - v0^2) (v0 = 0 since it is released from rest) so vf = sqrt(2 * a * d) d would be the 8 meters it slid, a would be the acceleration you found from the net force 2) Using conservation of energy. The only tricky part here is that you have to subtract the work done against the system due to friction. For constant force W = F*d so calculate potential difference from top of roof to 8 feet worth of sliding: PE = mgh h = d * sin(theta) (using trig) so this is how much energy is put into kinetic energy due to gravity. Now calculate energy taken away from system due to friction: again W = F * d F = mu * Fnormal (same as above) Ffr = .14 * m * g * cos(theta) W = Ffr * d then total energy gained: E = PE - W = mgh - Ffr*d this will ALL go to kinetic energy so set this value equal to (1/2)mv^2 (incidentally you end up with the exact same eq. as above): m * g * d * sin(theta) - mu * m * g * cos(theta) = (1/2) m * v^2 v^2 = 2 * (g * d * sin(theta) - mu * g * cos(theta)) take sqrt and you have the answer (notice that the mass cancels out due to normal force in friction eq.)
University Physics 1 Question:?
The roof of a two-story house makes an angle of 28° with the horizontal. A ball rolling down the roof rolls off the edge at a speed of 4.4 m/s. The distance to the ground from that point is 7.2 m. (a) How long is the ball in the air? 1 s (b) How far from the base of the house does it land? 2 m (c) What is its velocity just before landing? (Let upward be the positive y-direction.) x-component m/s ? y-component m/s ?
Physics, rolling + energy?
The problem statement has a problem. If the cylinder slides down it does not roll down. If id slides down one can treat it as a body of mass (m) however when id rolls down we have to deal with a rolling mass and consider moment of inertia (I) Where I=0.5 m R^2 also kinetic energy of the cylinder is Ke=0.5 I w^2 w - is angular velocity. And since energy is energy the energy at the top of the ramp (Pe) equal to kinetic energy (Ke) at the bottom. Ke=Pe=mgh (a) .5 I w^2=mgh w=sqrt(2mgh/I) w=sqrt(2mgh/.5 m R^2) w=sqrt(4gh/ R^2) h=Lsin(30)=3m w=sqrt(4 x 9.81 x 3/(.07)^2) w=155 rad/sec (b) In this case the angular speed has to be decomposed into vertical and horizontal components. The horizontal component will be responsible for motion away from the roof (and therefore distance) while the time in the air will be determined by the cylinders vertical movement. I’m assuming that the cylinder had no normal (to the cylinder) linear velocity and that only tangential component was present. In general angular speed (w) can be expressed w=V sin (angle between normal and tangential component) /R V- Linear component of the velocity (the outside surface of the cylinder) R- Radius of the cylinder Since the angle is assumed to be 90 degrees (no normal component) w=V/R V=w R Let’s compute vertical and horizontal components Vh=Vcos(30)=w R cos (30) Vv=Vsin(30) = 0.5 w R Time it takes fro the cylinder to hit the ground at freefall + initial velocity Vv H=Vvt +.5 gt^2 we have 5=(0.5 w R)t + .5 gt^2 Solve for t Then use that value (positive value) to compute the distance in question S=Vh t Let me know if you have any further questions.
Physics ball thrown off roof help please!?
To find the velocity required for an object to reach a certain height, calculate the velocity it would achieve from falling from that height. t=sqrt(2y/g) y=18.4m g=9.81m/s^2 t=1.94s (time to freefall 18.4m) assuming ball was launched from the ground Vy=V(original)+A(original)*t Vy=9.81*1.94s Vy=19.00m/s ___________________________________ b) time to reach maximum height is the same as the time it would take to drop from that height. t=1.94s ___________________________________ c) follow steps in a) to find how long it would take to drop 3.4m ___________________________________ d) Assuming it landed on the absolute top corner of the roof (can't see diagram). add times from a and c, and then divide 7 meters by that time.
Physics: Roof Top Problem?
Fix origin at the point from where the ball is thrown. Take horizontal towards the building as X axis and vertically upward as Y axis. a) In y direction:- Initial velocity = vyi Acceleration ay = - g = -9.8 m/s^2 Displacement Y = h1+h2 = 11 + 3.4 = 14.4 m Final velocity vyf = 0 (vertical velocity at max height = 0) vyf^2 = vyi^2 + 2 * ay * Y 0 = vyi^2 + 2 * (-9.8) * 14.4 0 = vyi^2 - 282.24 vyi = sqrt(282.24) = 16.8 m/s Ans: 16.8 m/s b) Let time taken = t vyf = vyi + ay * t 0 = 16.8 + (-9.8) * t 9.8 * t = 16.8 t = 16.8/9.8 = 1.7 s Ans: 1.7 s c) Time taken = sqrt(2*h2/g) = sqrt(2 * 3.4/9.8) = 0.83 s d) Let v = speed of throw v^2 = 2g*(h1+h2) v^2 = 2*9.8 * (11+3.4) = 282.24 m^2/s^2 Let initial horizontal velocity. v^2 = vxi^2 + vyi^2 vxi = sqrt(v^2 - vyi^2) = sqrt(282.24 - 16.8^2) = sqrt(282.24 - 282.44) = 0 Ans: 0
What are all the physical phenomenons responsible in rolling a ball?
Rolling a ball is an interesting process.For a perfect spherical ball and a surface , there will be only one point of contact at all times.When a force causes a push or a pull, the first reaction happens at the contact point.Being a rigid body , the force distributes itself to all the points on the ballThe point of contact being under the influence of friction has the least tendency to undergo motion when compared to other points on its surface. Hence the force shifts the rest of the mass about this point along the direction of push or pull. There are 2 parts to the velocity , one being angular velocity and another component being the linear velocity of the centre of gravity . Depending on the frictional force the ball either slips or rolls on the surface.
What is the kinetic energy of a solid sphere rolling down an inclined plane?
The sphere has two kings of kinetic energy: linear and rotational. The linear kinetic energy is given by the usual equation: 1/2 m*v^2. The rotational energy, in turna is given by 1/2*moment of inertia (I)* angular velocity^2 (w)^2. In rotational motion the moment of inertia, just like mass, is a measure of resistance to a change of motion; in this case a change in rotational motion. But the moment of inertia depends on the mass and the way the mass is distributed. For this reason different objects may have the same mass but differents moments of inertia, For a solid sphere I = 2/5*m*r^2, where r is the radius of the sphere. the last term of the rotational kinetic energy is the angular velocity , which may be expressed in radians/second or other units like the one shown below:linear velocityunits/linear units. In these last case you have to change the angular velocity to linear velocity.
PHYSICS HELP!!!!!!! The roof of a two-story house makes an angle of 27 degrees with the horizontal.?
It's kind of hard to explain w/o diagrams, but I'll give it a shot. To find time it is in the air, you need to know distance (d), acceleration (a), and initial velocity on the y-axis(Viy). You know d=7.4m, a=9.8m/s^2 (due to gravity). You can find Viy by thinking of 4.4 m/s as the hypotenuse of a right triangle. Viy is the length of the leg opposite the 23 degree angle. The sine of an angle is equal to the side opposite that angle divided by the hypotenuse, so sin(23)*4.4 m/s = Viy = 1.7 m/s (about). Now you can find the time it fell. I will start by also finding the final velocity (Vfy) using the equation Vf^2=Vi^2 + 2ad. Substitute in what you know and you get Vf^2=(1.7 m/s)^2 + 2*(9.8m/s^2)*7.4m. This gets you an answer of Vf=12 m/s. From here you can use the equation Vf = Vi + at. 12 m/s = 1.7 m/s + (9.8 m/s^2)*t. Solve this and you get t=1.1s. Finding Vf also answers the y component portion of part c. I will assume that the air resistance on the ball is negligible, so the velocity in the x (Vx) will be constant. This means you can use the equation d=vt to find the distance. You know t from part a. To find Vx you can use the same procedure as with Viy, except use cosine instead of sine. By doing this I found that Vx = 4.1 m/s (note: you should check all my math). Now back to the equation. d= 4.1 m/s * 1.1s. d= 4.5 m. Vx is the second part of the answer to part c, and d is the answer to part b. Ahh, woops. I used 23 degrees by mistake. I don't feel like fixing it, but it's correct except for that.
How would I calculate the acceleration of a ball rolling down a slope?
You would model the ball and slope to a degree of detail that you want, and then solve the equations of motion for that model.For instance, for rolling without slipping, and zero rolling resistance, you could use direct conservation of energy and understanding that the acceleration would be constant.[math]\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh[/math] … solve for v, the change in velocity and use kinematic equations and details of the slope to get the acceleration.There are many other approaches … ie using Lagrange mechanics methods… that may be more useful depending on the exact situation.