Answer:
Explanation:
Image of distant object will be made at far point or at 52.5 so
object distance u = infinity
image distance v = - 52.5 cm
focal length required = f
Lens formula
1 / v - 1 / u = 1 / f
1 / - 52.5 - 0 = 1 / f
f = -52.5 cm
= -.525 m
Power P = 1 / f = - 1 / .525
= - 1.90
now , for eye with glass we shall find new near point .
v = ?
u = - 17.2 cm
f = - 52.5 cm
1 / v - 1 / u = 1 / f
1 / v + 1 / 17.2 = - 1 / 52.5
1 / v = - 1 / 17.2 - 1 / 52.5
= - .05813 - .019
= - .07713
u = - 12.96 cm
so new near point will be 12.96 cm
Answer:
D. "The net force is zero, so the acceleration is zero"
Explanation:
edge 2020
Answer:
Two possible points
<em>x= 0.67 cm to the right of q1</em>
<em>x= 2 cm to the left of q1</em>
Explanation:
<u>Electrostatic Forces</u>
If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude

We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.





Equating


Operating and simplifying

To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.

Assuming the positive sign
:




Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative



The negative sign of x means q3 is located to the left of q1 (assumed in the origin).
Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North
Explanation:
In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).
Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC
AC2 = AB2 + BC2 ; AC^2 = 752 + 1552 ; from this we get AC = 172 km (3 significant figures)
Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees
Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North
Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North