Answer:
a. 2v₀/a b. 2v₀/a
Explanation:
a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.
Since the dragster starts from rest with an acceleration, a, using
s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster
s' = 0t + 1/2at²
s' = 1/2at²
Since the distance moved by me and the dragster must be the same,
s = s'
v₀t. = 1/2at²
v₀t. - 1/2at² = 0
t(v₀ - 1/2at) = 0
t= 0 or v₀ - 1/2at = 0
t= 0 or v₀ = 1/2at
t= 0 or t = 2v₀/a
So the maximum time tmax = 2v₀/a
b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s = v₀(2v₀/a)
= 2v₀/a
Answer:
1. Largest force: C; smallest force: B; 2. ratio = 9:1
Explanation:
The formula for the force exerted between two charges is

where K is the Coulomb constant.
q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.
For simplicity, let's combine Kq₁q₂ into a single constant, k.
Then, we can write

1. Net force on each particle
Let's
- Call the distance between adjacent charges d.
- Remember that like charges repel and unlike charges attract.
Define forces exerted to the right as positive and those to the left as negative.
(a) Force on A

(b) Force on B

(C) Force on C

(d) Force on D

(e) Relative net forces
In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

2. Ratio of largest force to smallest

Answer:
-10.8m/s^2
Explanation:
a=change in velocity/change in time
-27 m/s/2.5=10.8m/s^2
or if its not negative
27m/s/2.5=10.8m/s^2
Answer:
1.
Explanation:
Hello!
In this case, for such mathematical operations, we can wee that the slash represents a fraction or a division, say 8 ÷ 4 = 2, 6 ÷ 3 = 2, 20 ÷ 4 = 5, etc. In such a way, since the operation 2/2, represents 2 ÷ 2, it is clear that two is once in 2, therefore, the result is:
2 ÷ 2 = 1.
Best regards!