The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.
<h3>
Acceleration of the car </h3>
The acceleration of the car before stopping at the given distance is calculated as follows;
v² = u² + 2as
when the car stops, v = 0
0 = u² + 2as
0 = 15² + 2(76.5)a
0 = 225 + 153a
-a = 225/153
a = - 1.47 m/s²
<h3>Distance traveled when the speed is 32 m/s</h3>
If the same force is applied, then acceleration is constant.
v² = u² + 2as
0 = 32² + 2(-1.47)s
2.94s = 1024
s = 348.3 m
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Answer:

at t = 0.001 we have

at t = 0.01

at t = infinity

Explanation:
As we know that they are in series so the voltage across all three will be sum of all individual voltages
so it is given as

now we will have

now we have

So we will have

at t = 0 we have
q = 0

also we know that
at t = 0 i = 0




so we have

at t = 0.001 we have

at t = 0.01

at t = infinity

Answer:
11, 760 Pa.
Explanation:
By applying formula P= pgh, where P is pressure, p is density, g is gravitational acceleration (9.8 m\s2) and h is height of water level. Putting values in the formula, you can have the correct answer.
I hope it helps! good luck in chem!