A projectile motion is characterized by motion moving in a direction of an arc. It is acted upon by two component vectors: the horizontal and vertical. These two vectors are independent of each other when it comes to time of flight. The horizontal direction travels at constant speed, while the vertical direction travels at constant acceleration due to gravity, The time for an object to reach the ground would be equal, whether dropped from the sampe point or thrown in a projectile motion. Of course, this is assuming ideality wherein there is no air resistance.
So, the hang up time, or the time the object stayed on air is calculated using this equation:
a = Δv/t
Δv is the change in velocity which is the initial velocity when it was dropped to when it reaches zero velocity when it hits the ground.
9.81 m/s² = |(0 - 7.3)|/t
t = 0.744 seconds
Answer:
I would say the answer is A... but I'm not so sure ....
Answer:
the answer the correct one is c
Explanation:
Electric charges of different signs attract and those of the same sign repel. In addition, there are two types of insulating bodies, where the loads are fixed (immobile) and metallic (with mobile loads.
Let's analyze the situation presented
* A rod with positive approaches and the sphere is attracted, so the charge on the sphere is negative
* A rod with a negative charge approaches and the sphere is attracted, therefore the charge of the sphere must be positive.
For this to happen, the sphere must be unloaded and the charge that creates the phenomenon are induced charges because the mobile charges of the same sign as the sphere are repelled.
when checking the answer the correct one is c
Answer:
A = [kg]
B = [m/s²]
Explanation:
E = ½ Av² + Bmx
Substitute the units:
[J] = ½ A [m/s]² + B [kg] [m]
A Joule written in base units is:
1 J = 1 Nm = 1 kg m²/s²
Each term must have the same units.
[kg m²/s²] = A [m/s]²
[kg m²/s²] = A [m²/s²]
A = [kg]
[kg m²/s²] = B [kg] [m]
B = [m/s²]
