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Complete Question
Three equal point charges are held in place as shown in the figure below
If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.
A) F1=2F2
B) F1=3F2
C) F1=4F2
D) F1=9F2
Answer:
D) F1=9F2
Explanation:
We are told in the question that there are three equal point charges.
q, Q1, Q2 ,
q = Q1 = Q2
From the diagram we see the distance between the points d
q to Q1 = d
Q1 to Q2 = nd
Assuming n = 2
= 2 × d = 2d
Sum of the two distances = d + 2d = 3d
F1 is the force on q due to Q1 and
F2 is the force on q due to Q2,
Since we have 3 equal point charges and a total sum of distance which is 3d
Hence,
F1 = 9F2
Answer:
0.106
Explanation:
For 1 liter of diesel the car can get 19 km, if it takes 0.2 MJ for each km then it would take the total energy of 19*0.2 = 3.8 MJ to move an aerodynamic car 19 km. Since 1 liter of of diesel also contains 36 MJ in internal energy, then the efficiency of the diesel engine is the ratio of its output energy over its input energy:
Explanation:it is beause they are sharper and also have less surface area and therefore more pressure
<span>14 m/s
Assuming that all of the energy stored in the spring is transferred to dart, we have 2 equations to take into consideration.
1. How much energy is stored in the spring?
2. How fast will the dart travel with that amount of energy.
As for the energy stored, that's a simple matter of multiplication. So:
20 N * 0.05 m = 1 Nm = 1 J
For the second part, the energy of a moving object is expressed as
KE = 0.5 mv^2
where
KE = Kinetic energy
m = mass
v = velocity
Since we now know the energy (in Joules) and mass of the dart, we can substitute the known values and solve for v. So
KE = 0.5 mv^2
1 J = 0.5 0.010 kg * v^2
1 kg*m^2/s^2 = 0.005 kg * v^2
200 m^2/s^2 = v^2
14.14213562 m/s = v
So the dart will have a velocity of 14 m/s after rounding to 2 significant figures.</span>
Velocity =
(distance between start point and end point, regardless of the route traveled) / (time spent traveling).
That distance (called the "displacement"), is 10 meters, and almost exactly 1 hour is almost exactly 3,600 seconds. So the numerical value of the velocity during that time is
(10) / (3,600) = almost exactly 0.00278 m/s
= 2.78 x 10^-3 m/s.
