A muscle cramp is a sudden, painful contraction of a muscle, during which movement is difficult. What is happening at the cellular level during a muscle cramp?
OA
The proteins in the myofibrils become denatured, preventing relaxation.
B.
Myosin and actin do not fully detach, so the muscle cannot relax.
O C.
The motor neuron does not send a signal, so the muscle cannot relax.
OD
A lack of calcium causes the sarcomeres to lock together, preventing relaxation.

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).

The 10-kg double wheel with radius of gyration of 125mm about o is connected to the spring of stiffness k = 600 n/m by a cord which is wrapped securely around the inner hub. if the wheel is released from rest on the incline with the spring stretched 225 mm, calculate the maximum velocity v of its center o during the ensuing motion. the wheel rolls without slipping

A5 foot by 5 foot box culvert is buried 6 feet beneath the surface of the ground. a crosssection of this very long culvert (into and out of the page) is illustrated below. assume that the soil in which the culvert is buried has the following properties: jt = 120 pcf, w = 12%, ko = 0.50, ka = 0.33, and kp = 3.00. calculate the total horizontal force per unit length of culvert (fh) felt on one of the vertical faces of the culvert.