Answer: A flower pot falling
Explanation:
Answer:
1.F = 256 N
2.a = 8 m/s/s
By looking at the given information, you know force, and an acceleration. Therefore you have enough information to use the first formula.
F = ma
256 N = m * 8 m/s/s
m = 256 N/8 m/s/s
m = 32 Kg
If you are stationary, but in/on a moving vehicle/object you can be at rest and moving at then same time.
<u>Explanation</u>:
- A particle, when viewed from a given frame of reference, cannot be both at rest and in motion. However, in one frame of reference, a particle can be in motion whereas in another frame of reference the particle is in motion.
- For example, if you are seated in a plane, the plane is stationary in that reference frame and the Earth moves under it, but in the reference frame of the Earth, the plane is moving concerning the Earth. When you are standing still on Earth, in your frame of reference, the Earth is stationary, and the Sun and stars move around the Earth.
- However, in the frame of reference of the center of our solar system, the Earth orbits the Sun and the Sun are perturb slightly by the rest of the planets, but the rest of the galaxy orbits our solar system. Of course, in rest from our Galaxy, our solar system orbits a giant black hole at its center.
Answer:
FC vector representation
Magnitude of FC
Vector direction FC
degrees: angle that forms FC with the horizontal
Explanation:
Conceptual analysis
Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.
The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.
degrees
To calculate the magnitudes of the forces we apply Coulomb's law:
Equation (1): Magnitude of the electric force of the charge qA over the charge qC
Equation (2)
: Magnitude of the electric force of the charge qB over the charge qC
Known data
Problem development
In the equations (1) and (2) to calculate FAC Y FBC:
Components of the FBC force at x and y:
Components of the resulting force acting on qC:
FC vector representation
Magnitude of FC
Vector direction FC
degrees: angle that forms FC with the horizontal
Answer: w = 200N
Explanation:
w = mg
m = 20kg
g = 10m/s^ (approximately)
w = 20kg . 10m/s^
w = 200N
