Carbon is found in the solid form in geosphere of our earth. Coal and oil are some of the examples of materials containing carbon in the geosphere. when the coal or oil is burnt, carbon dioxide is formed and released in atmosphere. This carbon dioxide is absorbed by the water of the hydrosphere with the help of algae and plankton. The water turns acidic in nature. This way carbon is transferred from geosphere to hydrosphere.
Explanation:
For the equilibrium:
\rho_{wood}gh-\rho_{oil}g(h-x)-\rho_{water}gx=0ρ
wood
gh−ρ
oil
g(h−x)−ρ
water
gx=0
\rho_{wood}h-\rho_{oil}(h-x)-\rho_{water}x=0ρ
wood
h−ρ
oil
(h−x)−ρ
water
x=0
(974)(3.97)-928(3.97-x)-1000x=0(974)(3.97)−928(3.97−x)−1000x=0
x=2.54\ cmx=2.54 cm
I am not entirely sure but i believe the answer is C scupture
Answer: 1 is phone 2 is sandwich, Last is picture.
Explanation: I hoped That Helped !!
Answer:

Explanation:
First, we calculate the work done by this force after the box traveled 14 m, which is given by:
![W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m](https://tex.z-dn.net/?f=W%3D%5Cint%5Climits%5E%7Bx_f%7D_%7Bx_0%7D%20%7BF%28x%29%7D%20%5C%2C%20dx%20%5C%5CW%3D%5Cint%5Climits%5E%7B14%7D_%7B0%7D%20%28%7B18N-0.530%5Cfrac%7BN%7D%7Bm%7Dx%7D%29%20%5C%2C%20dx%5C%5CW%3D%5B%2818N%29x-%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7Bx%5E2%7D%7B2%7D%5D%5E%7B14%7D_%7B0%7D%5C%5CW%3D%2818N%2914m-%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7B%2814m%29%5E2%7D%7B2%7D-%2818N%290%2B%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7B0%5E2%7D%7B2%7D%5C%5CW%3D252N%5Ccdot%20m-52N%5Ccdot%20m%5C%5CW%3D200N%5Ccdot%20m)
Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

The box is initially at rest, so
. Solving for
:
