-- 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.
Answer:
We know that force applied per unit area is called pressure.
Pressure = Force/ Area
When force is constant than pressure is inversely proportional to area.
1- Calculating the area of three face:
A1 = 20m x 10 m =200 Square meter
A2 = 10 mx 5 m = 50 Square meter
A3 = 20m x 5 m = 100 Square meter
Therefore A1 is maximum and A2 is minimum.
2- Calculate pressure:
P = F/ A1 = 30 / 200 = 0.15 Nm⁻² ( minimum pressure)
P = F / A2 = 30 / 50 = 0.6 Nm⁻² ( maximum pressure)
Hence greater the area less will be the pressure and vice versa.
Answer:
Explanation:
As we stir the pot of soup by a metal spoon which is a conductor, it conducts the heat and starts heating after some time the temperature of the spoon rises and we are not able to hold the spoon and we get burns.
Answer:
The maximum speed of the car at the bottom of that drop is 26.34 m/s.
Explanation:
Given that,
The maximum vertical distance covered by the roller coaster, h = 35.4 m
We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :
v = 26.34 m/s
So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.
Answer:
2 x 10⁻³ volts
Explanation:
B = magnetic of magnetic field parallel to the axis of loop = 1 T
= rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²
θ = Angle of the magnetic field with the area vector = 0
E = emf induced in the loop
Induced emf is given as
E = B
E = (1) (20 x 10⁻⁴ )
E = 2 x 10⁻³ volts
E = 2 mV
