Answer:
Explanation:
Moment of inertia of the metal rod pivoted in the middle
= M l² / 12
If the spring is compressed by small distance x twisting the rod by angle θ
restoring force by spring
= k x
moment of torque about axis
= k x l /2
= k θ( l /2 )² ( x / .5 l = θ )
=
moment of torque = moment of inertia of rod x angular acceleration
k θ( l /2 )² = M l² / 12 d²θ/dt²
d²θ/dt² = 3 k/M θ
acceleration = ω² θ
ω² = 3 k/M
ω = √ 3 k / M
He preformed the first ever successful open heart surgery
Answer:
1. energy lost in the lever due to friction
3. visual estimation of height of the beanbag
5. position of the fulcrum for the lever affecting transfer of energy
Explanation:
Edge 2021
Answer:
a) 4500 cycles b) 0.0667s c) 6.67s
Explanation:
a) 15 Hz= 15 cycles/ s
5 mins= 300s
15 cycles/s * 300s= 4500 cycles
b) Period= 1/ frequency
Period= 1/ 15 cycles/s
Period= 0.0667s
c) Period * number of revolutions= time
0.0667 * 100= 6.67s
Answer:
0.572
Explanation:
First examine the force of friction at the slipping point where Ff = µsFN = µsmg.
the mass of the car is unknown,
The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.
First the tangential direction
∑Ft =Fft =mat
And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r
Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2
So going backwards and plugging in Ffc =m2atπr/ 2r =πmat
Ff = √(F2ft +F2fc)= matp √(1+π²)
µs = Ff /mg = at /g √(1+π²)=
1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572