Answer:
speed for last lap is 247.89 km/h
Explanation:
given data
velocity v1 = 203 km/h
velocity v2 = 199 km/h
no of lap n = 10
to find out
average speed for last lap
solution
we consider here distance d for 1 lap
so in first 9 lap time taken is
t1 = distance / velocity v2
t1 = 9d / 199 ...............1
and
for 10 lap time is
t2 = 10d / 203 .............2
so from 1 and 2 equation time for last lap
last lap time t3 = t2 - t1
t3 = 10d / 203 - 9d / 199
t3 = 0.004034 d
so speed for last lap is
speed = distance / time
speed = d / 0.004034 d
speed = 247.89 km/h
so speed for last lap is 247.89 km/h
Answer: 1.04N
Explanation:
Given
q1 = 2*10^-6C
q2 = 3.6*10^-6C
r = 0.25m
k = 9*10^9
Magnitude of electrostatic force can be calculated by using coulomb's law. Coulomb's law states that, "the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them."
F =(kq1q2) / r²
F = (9*10^9 * 2*10^-6 * 3.6*10^-6) / 0.25²
F = 0.0648/0.0625
F = 1.04N
The type of electrostatic force between the charges is the repulsive force
Answer:
20
To10
Explanation:
Which statement gives the gradient of this gr
<em><u>D. acceleration Hope this helps!</u></em>
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