<span>There are two possible arrangements of q1,q2,and q3 in this problem. They are:
q3, 2 cm gap, q1, 2 cm gap, q2
or
q1, 2/3 cm gap, q3, 4/3 cm gap, q2
We really don't care about the absolute magnitude of q, so the fact that it's 1.00 nano Coulombs is totally irrelevant to this problem. The only thing important is the relative charge and distances between the particles.
The force exerted between two particles is expressed as
F = q1*q2/r^2.
q1,q2 = charges on the particles.
r = distance between the particles.
Depending upon the relative charge (positive or negative) the force may be either attraction, or repulsion. But since the signs of all the charges mentioned are the same, I'll assume that the force will be repulsive.
For the distance between q1 and q3 I'll use the value "r". And since q1 and q2 are 2 cm apart, for the distance between q3 and q2, I'll use the value (2-r). So we have the following equations.
Force between q1 and q3
F = q1*q3/r^2
Force between q2 and q3
F = q2*q3/(2-r)^2
Set the 2 equations equal to each other
q1*q3/r^2 = q2*q3/(2-r)^2
Substitute the known values and solve for r.
q1*q3/r^2 = q2*q3/(2-r)^2
q*q/r^2 = 4q*q/(2-r)^2
q^2/r^2 = 4q^2/(4 - 4r + r^2)
1/r^2 = 4/(4 - 4r + r^2)
(4 - 4r + r^2)/(r^2(4 - 4r + r^2)) = 4/(4 - 4r + r^2)
(4 - 4r + r^2)/(r^2(4 - 4r + r^2)) = 4r^2/(r^2(4 - 4r + r^2))
0 = (4r^2-(4 - 4r + r^2))/(r^2(4 - 4r + r^2))
0 = (4r^2 - 4 + 4r - r^2)/(r^2(4 - 4r + r^2))
0 = (3r^2 - 4 + 4r)/(r^2(r-2)(r-2))
Now let's look at the numerator and denominator of the expression and see where we can get a value of 0. The denominator is allowed to have any value EXCEPT 0 and that will occur at r = 2, or r=0. And nothing else in the denominator will help the expression become 0. But if the numerator is 0, then the expression is 0. So let's see at what values the numerator is 0. Using the quadratic formula with A=3, B = 4 and C=-4, we get zeros at r = -2 and r = 2/3. Both of those values make sense. If r = -2, that means that the charges are arranges q3, q1, q2 with q1 being 2 cm from q1 and 4 cm from q2. And for r = 2/3, that also makes sense with the charges being arranged q1, q3, q2 and q3 is 2/3cm from q1 and 4/3cm from q2. In both cases, q3 is twice as from from q2 as it is from q1.</span>
<span>A lens is a piece of glass or other transparent substance with curved sides for concentrating or dispersing light rays, used singly (as in a magnifying glass) or with other lenses (as in a telescope). A binoculars, an eye, and a camera all contains a lens. A mirror does not contain a lens.</span>
Answer:
966.22 mph
Explanation:
Velocity of plane with respect to wind (Vp,w)= 612 mph east
velocity of wind with respect to ground, (Vw,g) = 362 mph at 15° North of
east
Write the velocities in vector form
Use the formula for the relative velocity
Where, V(p,w) is the velocity of plane with respect to wind
V(p,g) is the velocity of plane with respect to ground
V(w,g) is the velocity of wind with respect to ground
So,
Magnitude of velocity of lane with respect to ground
V(p,g) = 966.22 mph
Answer:
3.33 m/s²
Explanation:
If the acceleration is constant, it is equal to the average acceleration:
a = Δv / Δt
First, convert km/hr to m/s:
300 km/hr × (1 hr / 3600 s) × (1000 m/km) = 83.33 m/s
a = (83.33 m/s − 0 m/s) / 25 s
a = 3.33 m/s²
<h2>Option 1 is correct option .</h2>
Explanation:
The velocity of sound depends upon temperature . The temperature is different at different altitudes .
The velocity of sound has no effect of pressure . Thus at different altitudes , if pressure is different , it makes no difference .
Therefore the speed of sound varies rapidly with altitude , due to variation of temperature .
It also depends on the nature of medium . which also varies with altitude .
Moreover we cannot get distance by dividing time interval with 3 .