Answer:
The acceleration of the horse during this time interval is 24.286 m/min²
Explanation:
Given;
initial velocity of the horse during the gallop, u = 543 m /min
final velocity of the horse during the gallop, v = 628 m /min
time of motion, t = 3.5 minutes
The acceleration of the horse is given by the change in velocity per change in time;
Therefore, the acceleration of the horse during this time interval is 24.286 m/min²
Answer:
D. 66.4
Explanation:
So this problem uses SOHCAHTOA or the three trig functions.
Specifically this uses cosine, because it has an adjacent and a hypotenuse.
First you would determine what to do on the calculator, and since the problem is saying so, use the inverse cosine button. This will give you a angle measure from the decimal.
On a calculator, type in cos^-1(6/15). I used 6/15 because cosine is adjacent over hypotenuse. This will give you 66.4, which is D on the answers.
Answer:
the thickness of the film for destructive interference is 1 cm
Explanation:
We can assume that the radar wave penetrates the layer and is reflected in the inner part of it, giving rise to an interference phenomenon of the two reflected rays, we must be careful that the ray has a phase change when
* the wave passes from the air to the film with a higher refractive index
* the wavelength inside the film changes by the refractive index
λ = λ₀ / n
so the ratio for destructive interference is
2 n t = m λ
t = m λ / 2n
indicate that the wavelength λ = 2 cm, suppose that the interference occurs for m = 1, therefore it is thickness
t = 1 2/2 n
t = 1 / n
where n is the index of refraction of the anti-reflective layer. As they tell us not to take into account the change in wavelength when penetrating the film n = 1
t = 1 cm
So the thickness of the film for destructive interference is 1 cm
The order goes like this:
Radio, Micro waves, Infrared, Visible, UV (Ultra Violet), X-Ray, Gamma Ray.
