When someone lifts a book from the ground, the work you use is positive. By lifting the book, you change it's energy and it's original place The book gains, kinectic energy.
Hope I helped.
Answer:
66.4 N
Explanation:
From Newton's second law, <em>F </em>=<em> ma</em>
where <em>F</em> is the force, <em>m</em> is the mass and <em>a</em> is the acceleration.
Because the object has acceleration in two directions and the mass is constant, the force will be in two directions. The component of the forces are:
![F_x = ma_x = (7.00\text{ kg})(3.00 \text{ m/s}^2) = 21.0\text{ N}](https://tex.z-dn.net/?f=F_x%20%3D%20ma_x%20%3D%20%287.00%5Ctext%7B%20kg%7D%29%283.00%20%5Ctext%7B%20m%2Fs%7D%5E2%29%20%3D%2021.0%5Ctext%7B%20N%7D)
![F_y = ma_y = (7.00\text{ kg})(9.00 \text{ m/s}^2) = 63.0\text{ N}](https://tex.z-dn.net/?f=F_y%20%3D%20ma_y%20%3D%20%287.00%5Ctext%7B%20kg%7D%29%289.00%20%5Ctext%7B%20m%2Fs%7D%5E2%29%20%3D%2063.0%5Ctext%7B%20N%7D)
The magnitude of the resultant force is given by
![F = \sqrt{F_x^2+F_y^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Csqrt%7BF_x%5E2%2BF_y%5E2%7D)
![F = \sqrt{(21.0\text{ N})^2+(63.0\text{ N})^2} = \sqrt{(441.0\text{ N}^2)+(3969.0\text{ N}^2)} = \sqrt{(4410\text{ N}^2)} = 66.4 \text{ N}](https://tex.z-dn.net/?f=F%20%3D%20%5Csqrt%7B%2821.0%5Ctext%7B%20N%7D%29%5E2%2B%2863.0%5Ctext%7B%20N%7D%29%5E2%7D%20%3D%20%5Csqrt%7B%28441.0%5Ctext%7B%20N%7D%5E2%29%2B%283969.0%5Ctext%7B%20N%7D%5E2%29%7D%20%3D%20%5Csqrt%7B%284410%5Ctext%7B%20N%7D%5E2%29%7D%20%3D%2066.4%20%5Ctext%7B%20N%7D)
The correct answer should be C. standard
When you have standardization developed, then it is possible to compare scores across different scales and make sure that everything works relatively fine.
Answer:
![\Delta x = 3.65 \mu m](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%203.65%20%5Cmu%20m)
Explanation:
As we know that the sixth order maximum will have path difference given as
![\Delta x = N\lambda](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%20N%5Clambda)
here we know that
N = order of maximum
![\lambda = 609 nm](https://tex.z-dn.net/?f=%5Clambda%20%3D%20609%20nm)
now we have
![N = 6](https://tex.z-dn.net/?f=N%20%3D%206)
so we know that
![\Delta x = 6(609 nm)](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%206%28609%20nm%29)
![\Delta x = 3.65 \mu m](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%203.65%20%5Cmu%20m)
Energy cannot be created nor be destroyed