The minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.
<h3>
What is average speed?</h3>
The average speed of an object is the ratio of total distance traveled by the object to the total time of motion of the object.
<h3>Total time taken by the car during the entire race</h3>
time = distance/average speed
time = (1.41 km) / (278 km/h)
time = 0.0051 hr
The car travels the first half of the race, d (¹/₂ x 1410 m) at 210 km/h;
d = 705 m = 0.705 km
t1 = 0.705/210
t1 = 0.0034 hr
<h3>time for the second half</h3>
t2 = 0.0051 - 0.0034 hr
t2 = 0.0017 hr
<h3>minimum average speed of the second half</h3>
v = d/t
v = 0.705 km / 0.0017 hr
v = 414.7 km/hr
Thus, the minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.
Learn more about average speed here: brainly.com/question/4931057
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Answer:
a) E = 8628.23 N/C
b) E = 7489.785 N/C
Explanation:
a) Given
R = 5.00 cm = 0.05 m
Q = 3.00 nC = 3*10⁻⁹ C
ε₀ = 8.854*10⁻¹² C²/(N*m²)
r = 4.00 cm = 0.04 m
We can apply the equation
E = Qenc/(ε₀*A) (i)
where
Qenc = (Vr/V)*Q
If Vr = (4/3)*π*r³ and V = (4/3)*π*R³
Vr/V = ((4/3)*π*r³)/((4/3)*π*R³) = r³/R³
then
Qenc = (r³/R³)*Q = ((0.04 m)³/(0.05 m)³)*3*10⁻⁹ C = 1.536*10⁻⁹ C
We get A as follows
A = 4*π*r² = 4*π*(0.04 m)² = 0.02 m²
Using the equation (i)
E = (1.536*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.02 m²)
E = 8628.23 N/C
b) We apply the equation
E = Q/(ε₀*A) (ii)
where
r = 0.06 m
A = 4*π*r² = 4*π*(0.06 m)² = 0.045 m²
Using the equation (ii)
E = (3*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.045 m²)
E = 7489.785 N/C
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