Answer:
The input force that you use on an inclined plane is the force with which you push or pull an object. The output force is the force that you would need to lift the object without the inclined plane. This force is equal to the weight of the object.
Explanation:
 
        
             
        
        
        
They are incline hope this helps!
        
             
        
        
        
Answer:D
Explanation:Electric power=I*I*R
 =12*12*100
 =14400watts
 
        
             
        
        
        
Answer:
 K = -½U
Explanation:
From Newton's law of gravitation, the formula for gravitational potential energy is;
U = -GMm/R
Where, 
G is gravitational constant 
M and m are the two masses exerting the forces
R is the distance between the two objects
Now, in the question, we are given that kinetic energy is;
K = GMm/2R
Re-rranging, we have;
K = ½(GMm/R)
Comparing the equation of kinetic energy to that of potential energy, we can derive that gravitational kinetic energy can be expressed in terms of potential energy as;
K = -½U
 
        
             
        
        
        
Answer:
The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.
 Determine Fx."

Explanation:
We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation. 
torque=cross product of force and position . mathematically this can be express as 

Where 
 and the position vector
  and the position vector 

using the determinant method to expand the cross product in order to determine the torque we have 
![\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C2%26-3%262%5C%5C%20F_%7Bx%7D%20%267%26-5%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C)
by expanding we arrive at 

 since we have determine the vector value of the toque, we now compare with the torque value given in the question 

if we directly compare the j coordinate we have 
