10. A safe place to use the trebuchet would be away from other buildings and people. A good example of a place would be a large field with no nearby structures.
14. Many factors need to be kept consistent throughout the experiment. One example of a variable that would need to be consistent is the weight and size of the projectile.
15. It is important to do many trials so that you can make sure that the results of each trial are nearly the same. If they are all vastly different, then it means that something has gone wrong.
Sorry I was only able to answer a few questions, but I hope these few answers help! :)
Coulomb's Law
Given:
F = 3.0 x 10^-3 Newton
d = 6.0 x 10^2 meters
Q1 = 3.3x 10^-8 Coulombs
k = 9.0 x 10^9 Newton*m^2/Coulombs^2
Required:
Q2 =?
Formula:
F = k • Q1 • Q2 / d²
Solution:
So, to solve for Q2
Q2 = F • d²/ k • Q1
Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9
Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)
Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)
Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)
Then, take the reciprocal of the denominator and start
multiplying
Q2 = 1080 • 1 Coulombs/297
Q2 = 1080 Coulombs / 297
Q2 = 3.63636363636 Coulombs
Q2 = 3.64 Coulumbs
Answer:
The net force on the skater is zero. (
)
Explanation:
According to Newton's First Law, an object is at equilibrium when either it is at rest or moves at constant velocity, which means a net force of zero. Based on the given statement, there are no external forces acting on skate and, therefore, the net force on the skater is zero. (
)
Answer: 3.48g
Explanation:
here, we will be using conservation of momentum to solve the problem. i.e the total momentum remains unchanged, unless an external force acts on the system. We'll in thus question, there is no external force acting in the system.
Remember, momentum = mass * velocity, then
mass of blood * velocity of blood = combined mass of subject and pallet * velocity of subject and pallet
Velocity of blood = 56.5cm = 0.565m
mass of blood * 0.565 = 54kg * (0.000063/0.160)
mass of blood * 0.565 = 54 * 0.00039375
mass of blood * 0.565 = 0.001969
mass of blood = 0.00348kg
Thus, the mass of blood that leaves the heart is 3.48g