Answer:
Find two boxes just a bit smaller than the other. The long sides of the box should be less than twice as long as the short side of the boxes. The smaller box should fit inside the larger box with about 1 inch in each direction to spare. The boxes can be cut down so that they fit together properly. Leave the flaps on the boxes. Buy a small sheet of Plexiglas (tm) a little bit smaller than the width and length of the top of the box. You will also need four pieces of cardboard to use for reflectors.
Explanation:
To give solution to the exercise we must use the concepts of Torque, Vector magnitude and vector direction of the forces.
For the given problem we have to
In this way the torque acting on the particle as a function of distance and time is,
The net torque acting on the particle is
PART B) The direction of the torque is given by,
Therefore the torque direction is 48.04° below the x axis.
Answer:
When uranium is mined, it consists of approximately 99.3% uranium-238 (U238), 0.7% uranium-235 (U235), and < 0.01% uranium-234 (U234). These are the different uranium isotopes. Isotopes of uranium contain 92 protons in the atom's center or nucleus. (The number of protons in the nucleus is what makes the atoms "uranium.") The U238 atoms contain 146 neutrons, the U235 atoms contain 143 neutrons, and the U234 atoms contain only 142 neutrons. The total number of protons plus neutrons gives the atomic mass of each isotope — that is 238, 235, or 234, respectively. On an atomic level, the size and weight of these isotopes are slightly different. This implies that with the right equipment and under the right conditions, the isotopes can be separated.
Explanation:
Answer:
(A) The speed just as it left the ground is 30.25 m/s
(B) The maximum height of the rock is 46.69 m
Explanation:
Given;
weight of rock, w = mg = 20 N
speed of the rock at 14.8 m, u = 25 m/s
(a) Apply work energy theorem to find its speed just as it left the ground
work = Δ kinetic energy
F x d = ¹/₂mv² - ¹/₂mu²
mg x d = ¹/₂m(v² - u²)
g x d = ¹/₂(v² - u²)
gd = ¹/₂(v² - u²)
2gd = v² - u²
v² = 2gd + u²
v² = 2(9.8)(14.8) + (25)²
v² = 915.05
v = √915.05
v = 30.25 m/s
B) Use the work-energy theorem to find its maximum height
the initial velocity of the rock = 30.25 m/s
at maximum height, the final velocity = 0
- mg x H = ¹/₂mv² - ¹/₂mu²
- mg x H = ¹/₂m(0) - ¹/₂mu²
- mg x H = - ¹/₂mu²
2g x H = u²
H = u² / 2g
H = (30.25)² / 2(9.8)
H = 46.69 m
No person shall be judged by their financial status in the process of applying for a job or position of power in the government.
