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 
:
Explanation:
When the wire is connected to a battery, the compass needle moves and changes its position. This happens because the needle magnetizes the copper wire, thus, creating a force.
While the current in the wire produces a magnetic field and exerts a force on the needle. The insulation on the wire becomes energized and exerts a force on the needle. Hence, the compass needle moves and changes its position.
The average power output is the ratio between the work done to compress the spring, W, and the time taken, t:
(1)
The work done is equal to the elastic energy stored by the compressed spring:
where
is the spring constant and
is the compression of the spring. If we substitute the numbers, we find:
And now we can use eq.(1) to calculate the average power output:
Answer:
unknow e and f
Explanation:
In experiments with alpha particles that are obtained by the method of radioactive decay of atoms, some parameters are known
a) Known. The initial velocity is given by the energy of the particles entities by the atomic nuclei
b) Known. The particle charge always 2e, helium core
c) Known. It is set in the given experiment, in general it is selected as zero
d) Known. Placed by the experimenter
e) Unknown. The speed depends on the interactions with the system
f) Unknown. It depends on the interactions with the system, because the position depends on the interactions
g) Known. It is always the value of a helium nucleo
