In order to change the frictional force between two solid surfaces, it can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can.
Answer:
The x-component of the electric field at the origin = -11.74 N/C.
The y-component of the electric field at the origin = 97.41 N/C.
Explanation:
<u>Given:</u>
- Charge on first charged particle,
- Charge on the second charged particle,
- Position of the first charge =
- Position of the second charge =
The electric field at a point due to a charge 
at a point 
distance away is given by
where,
= Coulomb's constant, having value 
= position vector of the point where the electric field is to be found with respect to the position of the charge .
= unit vector along .
The electric field at the origin due to first charge is given by
is the position vector of the origin with respect to the position of the first charge.
Assuming, 
are the units vectors along x and y axes respectively.
Using these values,
The electric field at the origin due to the second charge is given by
is the position vector of the origin with respect to the position of the second charge.
Using these values,
The net electric field at the origin due to both the charges is given by
Thus,
x-component of the electric field at the origin = -11.74 N/C.
y-component of the electric field at the origin = 97.41 N/C.
Answer:
20.2 seconds
Explanation:
The airplane (and therefore the crate) initially has no vertical velocity, so v₀ = 0 m/s.
The crate is in free fall, so a = -9.8 m/s².
The crate falls downward, so Δx = -2000 m.
Find: t, the time it takes for the crate to land.
Δx = v₀ t + ½ at²
-2000 m = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 20.2 s
It takes 20.2 seconds for the crate to land.
80 km per hour i believe. i’ll admit i’m american so we don’t use km lol, but the math should be the same. total distance = 120km, and total time = 1.5 hours. 120/1.5 = 80.
