Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,

Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?

a = -1.47 m/
(a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/
(since same friction force is applied)

s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Answer:

Explanation:
First, let's find the voltage through the resistor using ohm's law:

AC power as function of time can be calculated as:
(1)
Where:

Because of the problem doesn't give us additional information, let's assume:

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

Answer:
8. acceleration =
= 1 unit .
9. acceleration =
= -1 unit.
10. acceleration =
= 0 units.
Explanation:
8. i) acceleration = velocity / time
ii) In this figure velocity = time
iii) therefore acceleration =
= 1 unit .
9. i) acceleration = velocity / time
ii) In this figure 4 = m + 5, therefore m = -1
therefore velocity = (-0.5
time) + 5
iii) therefore acceleration =
= -1 units.
10.) velocity is constant at 2
therefore acceleration =
= 0 units
the answer is D cuz electricity is a conductive