Answer:
When object is placed between the focus (F) and pole (P) of a concave mirror, magnified and erect image of the object is formed on the back of the mirror.
When object is placed between the centre of curvature and the principal focus of a concave mirror, magnified and inverted image is formed in front of the mirror.
Explanation:
Answer:
The width of the central bright fringe on the screen is observed to be unchanged is 
Explanation:
To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

Where,
w = width
wavelength
m is an integer, m = 1, 2, 3...
We here know that as
as w are constant, then

We need to find
, then

Replacing with our values:


Therefore the width of the central bright fringe on the screen is observed to be unchanged is 
The answer is A because at a young age children will not understand and they need to learn at the appropriate level
Answer:
maximumforce is F = mg
Explanation:
For this case we must use Newton's second law,
Σ F = m a
bold indicate vectors, so we will write it in its components x and y
X axis
Fₓ = maₓ
Axis y
Fy - W = m a
Now let's examine our case, with indicate that the bird is level, the force of the wings can have a measured angle with respect to the x axis, where the vertical component is responsible for the lift, let's use trigonometry to find the components
Cos θ = Fₓ / F
Fₓ = F cos θ
sin θ = Fy / F
Fy = F sin θ
Let's replace and calculate
F sin θ -w = m a
As the bird indicates that leveling at the same height, so the vertical acceleration is zero (ay = 0)
F sin θ = w = mg
The maximum value of this equation occurs when the sin=1, in this case
F = mg