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
:

Human are eurkaryotic. Eukaryotes are organisms that have cells with a nucleus. Prokayotes are cells without a nucleus. Protists are eukaryotic but they are usually single cellular. Also, human are not fungus. Fungus cannot move and have to get food from its surrounding area.
Answer:
Hi
Final temperature = 250.11 °C
Final volume = 0,1 m3.
Process work = 0
Explanation:
The specific volume in the initial state is: v = 0.1m3/2 kg = 0.05 m3/kg.
This volume is located between the volumes as saturated liquid and saturated steam at 20 °C. For this reason the water is initially in a liquid vapor mixture. As the piston was blocked the volume remains constant and the process is isometric, also known as isocoric process, so the final temperature will be the water temperature at a saturated steam of v=0.05m3/kg, which is obtained by using steam tables for water, by linear interpolation. As follows, using table A-4 of the Cengel book 7th Edition:
v=0.05 m3/kg
v1=0.057061 m3/kg
T1=242.56°C
v2=0.049779 m3/kg
T2=250.35°C
T=
The process work is zero because there is no change in volume during heating:
W=PxΔv=Px0=0
where
W=process work
P=pressure
Δv=change of volume, is zero because the piston was blocked so the volume remains constant.
r1 = 5*10^10 m , r2 = 6*10^12 m
v1 = 9*10^4 m/s
From conservation of energy
K1 +U1 = K2 +U2
0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2
0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2
M is mass of sun = 1.98*10^30 kg
G = 6.67*10^-11 N.m^2/kg^2
0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))
v2 = 5.35*10^4 m/s
Answer:
21s
Explanation:
Given parameters;
Radius = 10m
Speed or velocity = 3m/s
Unknown:
Period = ?
Solution:
To solve this problem, use the expression:
v =
r is the radius
T is the unknown
Input the parameters and solve for T;
3 =
62.84 = 3T
T = 21s