You're not going to like this answer, but it's the only one possible:. It wasn't I who learned anything in this unit. If it was either of us, it was YOU. I can't even tell from reading the question what the topic of the unit was. Was it pamphlets ? Microsoft Publisher ? Freshmen ? Getting Through High School ? This is a lot like asking me to write something "in your own words".
Answer:
λ = 451.7 nm
Explanation:
The expression for the constructive interference of the double diffraction experiment is
d sin θ = m λ
let's use trigonometry
tan θ = y / L
how the experiment occurs at very small angles
tan θ = sin θ / cos θ = sin θ
sin θ = y / L
we substitute
d y / L = m λ
λ = 
let's calculate
λ = 
λ = 4.51699 10⁻⁷ m
λ = 4.517 10⁻⁷ m (109 nm / 1m)
λ = 451.7 nm
Answer:
(a) convex mirror
(b) virtual and magnified
(c) 23.3 cm
Explanation:
The having mirror is convex mirror.
distance of object, u = - 20 cm
magnification, m = 1.4
(a) As the image is magnified and virtual , so the mirror is convex in nature.
(b) The image is virtual and magnified.
(c) Let the distance of image is v.
Use the formula of magnification.
Use the mirror equation, let the focal length is f.
Radius of curvature, R = 2 f = 2 x 11.67 = 23.3 cm
Answer:
30.63 m
Explanation:
From the question given above, the following data were obtained:
Total time (T) spent by the ball in air = 5 s
Maximum height (h) =.?
Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:
Total time (T) spent by the ball in air = 5 s
Time (t) taken to reach the maximum height =.?
T = 2t
5 = 2t
Divide both side by 2
t = 5/2
t = 2.5 s
Thus, the time (t) taken to reach the maximum height is 2.5 s
Finally, we shall determine the maximum height reached by the ball as follow:
Time (t) taken to reach the maximum height = 2.5 s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =.?
h = ½gt²
h = ½ × 9.8 × 2.5²
h = 4.9 × 6.25
h = 30.625 ≈ 30.63 m
Therefore, the maximum height reached by the cannon ball is 30.63 m
<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
