Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
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Answer:
0.76 mg/s
Explanation:
0.46 kg/week × (1 week / 7 days) × (1 day / 24 hrs) × (1 hr / 3600 s) × (1000 g/kg) × (1000 mg/g) = 0.76 mg/s
Answer:
Explanation:
Since both vectors are pointing on the same direction (Northeast), the sum of them will point in that same direction, and its magnitud will be the sum of the magnitudes of each vector (40m/s2+10m/s2). This problem is just a problem in one dimension. The sum of the vectors is then 50m/s2 Northeast.
B) 14.0 N
The way to solve this problem is to determine the kinetic energy the box had before and after the rough patch of floor. The equation for kinetic energy is:
E = 0.5 M V^2
where
E = Energy
M = Mass
V = velocity
Substituting the known values, let's calculate the before and after energy.
Before:
E = 0.5 M V^2
E = 0.5 13.5kg (2.25 m/s)^2
E = 6.75 kg 5.0625 m^2/s^2
E = 34.17188 kg*m^2/s^2 = 34.17188 joules
After:
E = 0.5 M V^2
E = 0.5 13.5kg (1.2 m/s)^2
E = 6.75 kg 1.44 m^2/s^2
E = 9.72 kg*m^2/s^2 = 9.72 Joules
So the box lost 34.17188 J - 9.72 J = 24.451875 J of energy over a distance of 1.75 meters. Let's calculate the loss per meter by dividing the loss by the distance.
24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N
Rounding to 1 decimal place gives 14.0 N which matches option "B".
