Answer:
During the segments B - C and D - E, the car stopped since the y axis is the distance and the distance stayed the same in between those segments.
For a simpler answer, the flat horizontal lines on the graph are the times when the car was stopped.
Answer:
410 m
Explanation:
Given:
v₀ = 20.5 m/s
a = 0 m/s²
t = 20 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (20.5 m/s) (20 s) + ½ (0 m/s²) (20 s)²
Δx = 410 m
Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )
Answer:
hi there
Explanation:
1 - III
2- 1
3-1
HOPE IT HELP YOU
PLz mark me as a BRAINLIST
Answer: search it on browser
