Answer:
The correct answer is letter b. less than
Explanation:
In the attached schematic figure, the lens (lens) is initially adjusted for a given distance from the object. If the object gets closer, the image loses its sharpness. To recover it, the lens settles down, increasing convergence, that is, decreasing the focal length.
Answer: 34.9 cm
Explanation:
You are given the following parameters;
Object distance U = 32 cm
Magnification M = - 12.0
According to formula for magnification;
M = V/U
Where V = image distance.
Substitute V and M into the formula
-12 = V/32
Cross multiply
V = -12 × 32
V = - 384
You can use the formula
1/f = 1/V + 1/U
Where f = focal length
Substitute V and U into the formula
1/f = - 1/384 + 1/32
Find the lowest common factor of the denominator at Left hand side
1/f = ( -1 + 12 ) / 384
1/f = 11/384
Reciprocate both sides
F = 384/11
F = 34.9 cm
He should therefore use the focal length of 34.9 cm
Answer:
Explanation:
From the data it appears that A is the middle point between two charges.
First of all we shall calculate the field at point A .
Field due to charge -Q ( 6e⁻ ) at A
= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴
= 13.82 x 10⁻⁶ N/C
Its direction will be towards Q⁻
Same field will be produced by Q⁺ charge . The direction will be away
from Q⁺ towards Q⁻ .
We shall add the field to get the resultant field .
= 2 x 13.82 x 10⁻⁶
= 27.64 x 10⁻⁶ N/C
Force on electron put at A
= charge x field
= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶
= 44.22 x 10⁻²⁵ N
If any of them do, it's not intentional, that is, it's not part of
the design of the communications satellite.
Communications satellites communicate with their customers
and their control centers on the ground using plain old UHF radio
and microwave signals.
