Answer:
a) They are in the same point
b) t = 0 s, t = 2.27 s, t = 5.73 s
c) t = 1 s, t = 4.33 s
d) t = 2.67 s
Explanation:
Given equations are:


Constants are:

a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both
and
depend on t and don't have constant terms.
So both cars A and B are in the same point.
b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).


s,
s
c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.

s,
s
d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.
s
Answer:
Option (3)
Explanation:
Formula used to calculate acceleration is,
F = ma
Where F = force exerted on a mass
m = mass
a = acceleration due to force exerted on the mass
Option (1),
When F = 100 N and m = 100 kg
100 = 100a
a = 1 m per sec²
Option (2)
For F = 1 N and m = 100 kg
1 = 100a
a = 
a = 0.01 m per sec²
Option (3)
For F = 100 N and m = 1 kg
100 = 1(a)
a = 100 m per sec²
Option (4)
For F = 1 N and m = 1 kg
1 = 1(a)
a = 1 m per sec²
Therefore. acceleration in Option (3) is the maximum.
Answer: 1100 W
Explanation:
Input power = 220(5) = 1100 W
The transformer will step up/down voltage, but will also step down/up current.
Neglecting hysteresis and other minor losses, the power will remain the same.
Answer:
F' = (4/9)F
Explanation:
The electrostatic force between two charged objects is given by Coulomb's Law:
F = kq₁q₂/r² -------------------- equation (1)
where,
F = Electrostatic Force
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of second charge
r = distance between charges
Now, when the charges and distance altered as follows:
q₁' = 2q₁
q₂' = 2q₂
r' = 3r
Then,
F' = kq₁'q₂'/r'²
F' = k(2q₁)(2q₂)/(3r)²
F' = (4/9)kq₁q₂/r²
using equation (1):
<u>F' = (4/9)F</u>
P = IV
I = P/V = 30 / 120 = 0.25 A.
Current = 0.25A