-- Class I lever
The fulcrum is between the effort and the load.
The Mechanical Advantage can be anything, more or less than 1 .
Example: a see-saw
-- Class II lever
The load is between the fulcrum and the effort.
The Mechanical Advantage is always greater than 1 .
Example: a nut-cracker, a garlic press
-- Class III lever
The effort is between the fulcrum and the load.
The Mechanical Advantage is always less than 1 .
I can't think of an example right now.
Your answer is -6 did that answer your question
Answer: The spring constant is K=392.4N/m
Explanation:
According to hook's law the applied force F will be directly proportional to the extension e produced provided the spring is not distorted
The force F=ke
Where k=spring constant
e= Extention produced
h=2m
Given that
e=20cm to meter 20/100= 0.2m
m=100g to kg m=100/1000= 0.1kg
But F=mg
Ignoring air resistance
assuming g=9.81m/s²
Since the compression causes the plastic ball to poses potential energy hence energy stored in the spring
E=1/2ke²=mgh
Substituting our values to find k
First we make k subject of formula
k=2mgh/e²
k=2*0.1*9.81*2/0.1²
K=3.921/0.01
K=392.4N/m
I think it was by boat and coming from over seas
Answer:
t = 5.56 ms
Explanation:
Given:-
- The current carried in, Iin = 1.000002 C
- The current carried out, Iout = 1.00000 C
- The radius of sphere, r = 10 cm
Find:-
How long would it take for the sphere to increase in potential by 1000 V?
Solution:-
- The net charge held by the isolated conducting sphere after (t) seconds would be:
qnet = (Iin - Iout)*t
qnet = t*(1.000002 - 1.00000) = 0.000002*t
- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:
V = k*qnet / r
Where, k = 8.99*10^9 ..... Coulomb's constant
qnet = V*r / k
t = 1000*0.1 / (8.99*10^9 * 0.000002)
t = 5.56 ms