Your pendulum does a complete swing in 1.9 seconds. You want to SLOW IT DOWN so it takes 2.0 seconds.
Longer pendulums swing slower.
You need to <em>make your pendulum slightly longer</em>.
If your pendulum is hanging by a thread or a thin string, then its speed doesn't depend at all on the weight at the bottom. You can add weight or cut some off, and it won't change the speed a bit.
Answer:
The answer is either 5 or I'm learning something different and I just can't read
Explanation:
I hope this helped...
net force on Maze during his sliding downwards motion
now we will have
now to find the acceleration we will have
now to find the speed after 4 s is given by kinematics
here we know that
so Maze speed after 4 s will be 16.22 m/s
Answer:
ventriculoperitoneal
Explanation:
A VP or ventriculoperitoneal shunt is a medical equipment or a device which is used to relieve pressure from the brain that is caused by the accumulation of fluid in the brain. Ventriculoperitoneal shunting is a medical procedure that is primarily used to treat a condition known as hydrocephalus. This condition occurs when the more cerebrospinal fluid (CSF) gets collected in the ventricles of the brain than required.
Hence the answer is ---
ventriculoperitoneal
Answer:
V(t) = (q0/C) * e^(−t/RC
)
Explanation:
If there were a battery in the circuit with EMF E , the equation for V(t) would be V(t)=E−(RC)(dV(t)/dt) . This differential equation is no longer homogeneous in V(t) (homogeneous means that if you multiply any solution by a constant it is still a solution). However, it can be solved simply by the substitution Vb(t)=V(t)−E . The effect of this substitution is to eliminate the E term and yield an equation for Vb(t) that is identical to the equation you solved for V(t) . If a battery is added, the initial condition is usually that the capacitor has zero charge at time t=0 . The solution under these conditions will look like V(t)=E(1−e−t/(RC)) . This solution implies that the voltage across the capacitor is zero at time t=0 (since the capacitor was uncharged then) and rises asymptotically to E (with the result that current essentially stops flowing through the circuit).
