Answer: 
Explanation:
Given
Length of plank is 1.6 m
Force
is applied on the left side of plank
Force
is applied 43 cm from the left end O.
Mass of the plank is 
for equilibrium
Net torque must be zero. Taking torque about left side of the plank

Net vertical force must be zero on the plank

Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

W = 3.266 N
The mass of the meters stick is :

So, the mass of the meter stick is 0.333 kg.
This question is wrong because in momontum we will write acceleration instead of speed. suppose acceleration is 5m/s2 then
P= ma
then put values
Explanation:
A) Use Hooke's law to find the spring constant.
F = kx
40 N = k (0.4 m)
k = 100 N/m
B) Period of a spring-mass system is:
T = 2π √(m / k)
T = 2π √(2.6 kg / 100 N/m)
T = 1 s
Frequency is the inverse of period.
f = 1 / T
f = 1 Hz