-- 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.
The ships should get suck together as the water must be replaced from the sides. Please mark Brainliest!!!
Given that force is applied at an angle of 30 degree below the horizontal
So let say force applied if F
now its two components are given as


Now the normal force on the block is given as



now the friction force on the cart is given as



now if cart moves with constant speed then net force on cart must be zero
so now we have




so the force must be 199.2 N
Answer:
2.11eV
Explanation:
We know that speed of light is it's wavelength times frequency.

Planck's constant is 
The energy gap is calculated by multyplying the light's frequency by planck's constant:

Hence, the energy gap is 2.11eV