The zero net electric field point is at a point that is 0.98 m away from 4.7C charge.If a 14C charge is placed at this point then, force acted on the charge placed at this point is equal to zero.
Explanation:
Let at A both net electric field is zero then
At A ,E1=E2
E1=k*Iq1I / (d+x)^2
E2=k*Iq2I /x^2
Equating both
Answer:
maximum speed of the bananas is 18.8183 m/s
Explanation:
Given data
amplitude A = 23.195 cm
spring constant K = 15.2676 N/m
mass of the bananas m = 56.9816 kg
to find out
maximum speed of the bananas
solution
we know that radial oscillation frequency formula that is = √(K/A)
radial oscillation frequency = √(15.2676/23.195)
radial oscillation frequency is 0.8113125 rad/s
so maximum speed of the bananas = radial oscillation frequency × amplitude
maximum speed of the bananas = 0.8113125 × 23.195
maximum speed of the bananas is 18.8183 m/s
Answer:
the last one, the third one, and the first one.
Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.