Answer:
Explanation:
The equation of equlibrium for the box is:
The formula for the acceleration, given in 
, is:
Velocity can be derived from the following definition of acceleration:
![v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }](https://tex.z-dn.net/?f=v%20%3D%5Csqrt%7B2%5Ccdot%5B%282.278%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%29%5Ccdot%20x%20%7C_%7B0%5C%2Cm%7D%5E%7B27%5C%2Cm%7D-%280.034%5C%2C%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%7D%29%5Ccdot%20x%5E%7B2%7D%7C_%7B0%5C%2Cm%7D%5E%7B27%5C%2Cm%7D%5D%20%20%7D)
The speed after the box has travelled 17 meters is:
Answer:
v = -14 m/s
Explanation:
Given that,
Initial location of the ball, X₁ = 10 m
Final position of the ball, X₂ = -25 m
Time taken to travel is, t = 2.5 s
The average velocity of the ball is given by the formula,
V = X₂ - X₁ / t m/s
Substituting the values in the above equation,
V = -25 - 10 / 2.5
= -14 m/s
The negative sign in the velocity indicates that ball rolls in the opposite direction.
Hence, the average velocity of the ball is v = -14 m/s
Answer:
8.1 x 10^13 electrons passed through the accelerator over 1.8 hours.
Explanation:
The total charge accumulated in 1.8 hours will be:
Total Charge = I x t = (-2.0 nC/s)(1.8 hrs)(3600 s/ 1 hr)
Total Charge = - 12960 nC = - 12.96 x 10^(-6) C
Since, the charge on one electron is e = - 1.6 x 10^(-19) C
Therefore, no. of electrons will be:
No. of electrons = Total Charge/Charge on one electron
No. of electrons = [- 12.96 x 10^(-6) C]/[- 1.6 x 10^(-19) C]
<u>No. of electrons = 8.1 x 10^13 electrons</u>
Answer:
a) U = 735 J
, b) U = 125.7 J
, c) U = 0 J
Explanation:
The gravitational power energy is
U = mg y - mg y₀
The last value is a constant, for simplicity we can make it zero, if the lowest point is at the origin of the coordinate system, which in this case we will place in the lowest part
a) Rope is horizontal
The height in this case is the same length of the rope
y = 2.10 m
w = mg = 350 N
U = 350 2.10
U = 735 J
b) when the angle is 34º
y = L - L cos 34
y = L (1- cos34)
y = 2.10 (1- cos 34)
y = 0.359 m
U = 350 0.359
U = 125.7 J
c) in this case this point coincides with the reference system
y = 0
U = 0 J
