The  force applied to lift the crate is 171 N
Explanation:
The lever works on the principle of equilibrium of moments, so we can write:

where
 is the  force in input
 is the  force in input
 is the arm of the input force
 is the arm of the input force
 is the output force
 is the output force
 is the arm of the output force
 is the arm of the output force
For the lever in this problem, we have:


 (force applied)
 (force applied)
Solving the equation for  , we find the force applied to lift the crate:
, we find the force applied to lift the crate:

Learn more about levers:
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Answer:
Choice A: approximately  , assuming that the two pistons are connected via some confined liquid to form a simple machine.
, assuming that the two pistons are connected via some confined liquid to form a simple machine. 
Explanation:
Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:
 .
.
By Pascal's Principle, because the first piston exerted a pressure of  on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.
 on the liquid, the liquid will now exert the same amount of pressure on the walls of the container. 
Assume that the second piston is part of that wall. The pressure on the second piston will also be  . In other words:
. In other words:
 .
.
To achieve a force of  , the surface area of the second piston should be:
, the surface area of the second piston should be:
 .
.
 
        
             
        
        
        
Answer:
I know that T= kx where T is the tension which equaka the force og gravity = mg = 1.37 * 10 = 13.7 x is the elongation of the spring so the length after dangling the object minus the original length.
I hope it helps 
plz let me know if it is wrong or right.
 
        
             
        
        
        
Answer:
The lose of thermal energy is, Q = 22500 J
Explanation:
Given data,
The mass of aluminium block, m = 1.0 kg
The initial temperature of block, T = 50° C
The final temperature of the block, T' = 25° C
The change in temperature, ΔT = 50° C - 25° C
                                                      = 25° C
The specific heat capacity of aluminium, c = 900  J/kg°C
The formula for thermal energy,
                              <em>Q = mcΔT</em>
                                  = 1.0 x 900 x 25
                                  = 22500 J
Hence, the lose of thermal energy is, Q = 22500 J
 
        
             
        
        
        
A non <span>foliated </span>rock has interlocking grains with no specific pattern.