Angular Momentum
... How do we show that A B ( Ay Bz Az By )iˆ ( Az Bx Ax Bz ) ˆj ( Ax By Ay Bx )kˆ ? ...
... How do we show that A B ( Ay Bz Az By )iˆ ( Az Bx Ax Bz ) ˆj ( Ax By Ay Bx )kˆ ? ...
5.1 Net Force
... To obey Newton’s First law of motion (inertia) then the weight is balanced by a force pointing away from the center of the Earth of the same magnitude called a tension on the balance. ...
... To obey Newton’s First law of motion (inertia) then the weight is balanced by a force pointing away from the center of the Earth of the same magnitude called a tension on the balance. ...
apPhysics_lec_06
... Gravity or Weight (Gravitational force Earth pulling on objects around it): W ...
... Gravity or Weight (Gravitational force Earth pulling on objects around it): W ...
Lab7_StaticEquilibrium
... on it is zero and thus the object is not accelerating. Of course, the word static implies that the object’s velocity is zero as well. This is fine for point particles, objects that do not have size. This is the approximation that we have been making so far in class. When you remove that approximatio ...
... on it is zero and thus the object is not accelerating. Of course, the word static implies that the object’s velocity is zero as well. This is fine for point particles, objects that do not have size. This is the approximation that we have been making so far in class. When you remove that approximatio ...
PROBLEMS ON MECHANICS
... ing string or rod undergoes a virtual This allows us to write down the conlengthening of ∆x, then T = (∆Π − dition of torque balance for the hanging portion of the rope (as we know the ho∑i δ⃗xi · ⃗Fi )/∆x. rizontal coordinate of its centre of mass). The method can also be used for find- The next pr ...
... ing string or rod undergoes a virtual This allows us to write down the conlengthening of ∆x, then T = (∆Π − dition of torque balance for the hanging portion of the rope (as we know the ho∑i δ⃗xi · ⃗Fi )/∆x. rizontal coordinate of its centre of mass). The method can also be used for find- The next pr ...
R - IBPhysicsLund
... Topic 2: Mechanics 2.4 Uniform circular motion 2.4.1 Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle. 2.4.2 Apply the expressions for centripetal acceleration. 2.4.3 Identify the force produci ...
... Topic 2: Mechanics 2.4 Uniform circular motion 2.4.1 Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle. 2.4.2 Apply the expressions for centripetal acceleration. 2.4.3 Identify the force produci ...
Oaks_Park - TuHS Physics Homepage
... the train for this!) Also calculate what the coefficient of friction has to be in order for the locomotive to do this. (The normal force would be due to the mass of only the locomotive.) (10 pts) D) Power Using the kinetic energy as work and the time from A), calculate the power output of the train ...
... the train for this!) Also calculate what the coefficient of friction has to be in order for the locomotive to do this. (The normal force would be due to the mass of only the locomotive.) (10 pts) D) Power Using the kinetic energy as work and the time from A), calculate the power output of the train ...
Ch 5 Newtons Laws of Motion
... • In this equation, µs is the coefficient of static friction. • In general, µs is greater than µk. This means that the force of static friction is usually greater than the force of kinetic friction. • Friction plays an important role in driving safety. • When a car is moving with its tires rolling f ...
... • In this equation, µs is the coefficient of static friction. • In general, µs is greater than µk. This means that the force of static friction is usually greater than the force of kinetic friction. • Friction plays an important role in driving safety. • When a car is moving with its tires rolling f ...
Vectors and Scalars * Learning Outcomes
... e.g. Find the vertical and horizontal components of a vector of magnitude 20 N acting at 60o to the horizontal. e.g. Michelle pulls a rope which is tied to a cart with a force 300 N. The rope makes an angle of 20o to the horizontal. Find the effective vertical and horizontal forces on the cart d ...
... e.g. Find the vertical and horizontal components of a vector of magnitude 20 N acting at 60o to the horizontal. e.g. Michelle pulls a rope which is tied to a cart with a force 300 N. The rope makes an angle of 20o to the horizontal. Find the effective vertical and horizontal forces on the cart d ...
Document
... 5.7.2. A ball on the end of a rope is moving in a vertical circle near the surface of the earth. Point A is at the top of the circle; C is at the bottom. Points B and D are exactly halfway between A and C. Which one of the following statements concerning the tension in the rope is true? a) The tens ...
... 5.7.2. A ball on the end of a rope is moving in a vertical circle near the surface of the earth. Point A is at the top of the circle; C is at the bottom. Points B and D are exactly halfway between A and C. Which one of the following statements concerning the tension in the rope is true? a) The tens ...
香港考試局
... descending pan comes into contact with the plate when the two pans are at the same level. The motion of the system becomes simple harmonic until it comes to rest momentarily. With the contact point taken as the origin, find the equilibrium position and the angular frequency of the motion. (Assume th ...
... descending pan comes into contact with the plate when the two pans are at the same level. The motion of the system becomes simple harmonic until it comes to rest momentarily. With the contact point taken as the origin, find the equilibrium position and the angular frequency of the motion. (Assume th ...
Polygon of Forces
... In some cases the vector diagram, Figure 19b, may close by itself, so that R = 0; however, this does not necessarily mean that the sum of the moments is also zero. Consider the special case of two equal and opposite forces, F, distance d apart, as shown in Figure 20. Clearly there is no resultant fo ...
... In some cases the vector diagram, Figure 19b, may close by itself, so that R = 0; however, this does not necessarily mean that the sum of the moments is also zero. Consider the special case of two equal and opposite forces, F, distance d apart, as shown in Figure 20. Clearly there is no resultant fo ...
Exam and Revision Advice
... Net force on m1 equals T, but T is unknown, so use Net force on the system of two masses: Net force on (m1 and m2) = m2g, but Net F = Ma, so m2g = (m1 + m2) x a. Substituting, 0.10 x 10 = (0.1 + 0.4) x a, solve for a. Conseq Q’n: use value of accel’n. Two methods: i) Use equations of motion to find ...
... Net force on m1 equals T, but T is unknown, so use Net force on the system of two masses: Net force on (m1 and m2) = m2g, but Net F = Ma, so m2g = (m1 + m2) x a. Substituting, 0.10 x 10 = (0.1 + 0.4) x a, solve for a. Conseq Q’n: use value of accel’n. Two methods: i) Use equations of motion to find ...